Does L^1 (seminorm) convergence imply a.e. convergence?

[Fredrik]

Define $\rho(f)=\int |f|\mathrm d\mu$ for all integrable $f:X\to\mathbb C$. This $\rho$ is a seminorm, not a norm. Does $\rho(f_n-f)\to 0$ imply $f_n\to f$ a.e.?

I kind of think that it should, because in the case of real-valued functions, $\int|f_n-f|\mathrm d\mu$ is the area between the graphs, and I don't see how it can go to zero without $f_n\to f$ a.e. I have a feeling that this should be easy, but I don't see how to prove it. Going to bed now. Maybe I'll see it immediately in the morning.

 

[micromass]

It's false, consider the following function for $0\leq k <2^n$:

$$f_{k,n}(x) = \left\{\begin{array}{l}1 ~\text{if}~x\in [k/2^n,(k+1)/2^n]\\ 0~\text{otherwise}\end{array}\right.$$

Then consider the sequence

$$f_{0,1},~f_{1,1},~f_{0,2},~f_{1,2},~f_{2,2},~f_{3,2},~f_{0,3},...$$

So you get smaller and smaller rectangles. You can see this sequence converges to $0$ in $L^1$ but not a.e.

However, you do know that if a sequence converges in $L^1$, then it converges in measure and thus for any subsequence (for example the sequence itself), there is a subsequence that converges a.e.

 

[homeomorphic]

It is a norm if you identify functions that agree a.e.

The counter-example is one where the sequence doesn't converge. I think it could be true if you assume in advance that f_n converges a.e. For example, if you had an integrable bound, you could use Lebesgue dominated convergence to show this.

I posted an argument for the general case, but wasn't sure about it, so I deleted it. I realized I'd probably have to make sure I was actually completely rigorous about it to avoid error in this case. I think I just showed convergence in measure, actually.

 

[micromass]

I think it could be true if you assume in advance that f_n converges a.e.

Certainly, if $f_n$ converges in $L^1$ to some function $f$ and a.e. to some function $g$, then it converges in measure to both $f$ and $g$. Since convergence in measure is unique (almost everywhere), we get f=g a.e and thus $f_n$ converges to $f$ a.e.

 

[Fredrik]

It's false, consider the following function for $0\leq k <2^n$:

$$f_{k,n}(x) = \left\{\begin{array}{l}1 ~\text{if}~x\in [k/2^n,(k+1)/2^n]\\ 0~\text{otherwise}\end{array}\right.$$

Then consider the sequence

$$f_{0,1},~f_{1,1},~f_{0,2},~f_{1,2},~f_{2,2},~f_{3,2},~f_{0,3},...$$

So you get smaller and smaller rectangles. You can see this sequence converges to $0$ in $L^1$ but not a.e.

Awesome. I can see it. Thank you.

However, you do know that if a sequence converges in $L^1$, then it converges in measure and thus for any subsequence (for example the sequence itself), there is a subsequence that converges a.e.

I know that thing about subsequences (I worked through the proof a couple of days ago), but not that $L^1$ convergence implies in measure convergence. Is it really true? Friedman's book (Foundations of modern analysis) claims to prove the following (Edit: No, it doesn't): A sequence $\langle f_n\rangle_{n=1}^\infty$ of integrable simple functions satisfies the statements

(a) $\langle f_n\rangle_{n=1}^\infty$ is $L^1$-Cauchy.
(b) $f_n\to f$ a.e.

if and only if it satisfies the statements

(a) $\langle f_n\rangle_{n=1}^\infty$ is $L^1$-Cauchy.
(b') $f_n\to f$ in measure.

I haven't studied that proof, but I will start as soon as I've posted this. $L^1$ convergence implies $L^1$ "Cauchyness". So if every $L^1$ convergent sequence is convergent in measure, then every $L^1$ convergent sequence of integrable simple functions satisfies (a) and (b') and should therefore satisfy (a) and (b), and in particular converge a.e.

Interesting...your example satisfies (a) and (b') but not (b), and it's a sequence of integrable simple functions, so it seems to directly contradict the theorem. Maybe Friedman's "theorem" is just wrong. I will keep that in mind as I study the proof (or "proof"). I'll post an update when I know more.

Edit: I see now that I had it wrong. The sequence that satisfies (a) and (b) is not the same sequence that satisfies (a) and (b'). The argument goes like this: If $\langle f_n\rangle_{n=1}^\infty$ is a sequence of integrable simple functions that satisfies (a) and (b)', then it's Cauchy in measure. This implies that it has a subsequence that converges almost uniformly. That's the sequence that satisfies (a) and (b).

Friedman also proves that every sequence of integrable simple functions that's $L^1$-Cauchy is convergent in measure.

 

[micromass]

That convergence in $L^1$ implies convergence in measure follows from Chebyshev's (also called Markov's) inequality:

$$\mu\{x~\vert~|f_n(x) - f(x)|\geq \varepsilon\} = \int_{\{x~\vert~|f_n(x) - f(x)|\geq \varepsilon\}} 1d\mu(x) \leq \frac{1}{\varepsilon} \int |f_n(x)-f(x)|d\mu(x)$$

You should check out Bartle's "Elements of Integration and Lebesgue Measure". It has an entire chapter on modes of convergence and its innerrelations.

 

[Fredrik]

Thanks, I'll check it out. I think I have everything I need for the moment sorted out, but I'm sure something new will pop up, and I think that neither Friedman nor Lang (my two sources for the approach to Lebesgue integration based on the idea $\int f\mathrm d\mu=\lim_n\int f_n\mathrm d\mu$ where the $f_n$ are integrable simple functions) covers this really well.

 

[Fredrik]

I still have a problem, because I don't want to use Chebyshev's inequality at this point. I think I need to explain what I'm trying to do. I'm trying to sort out some details in the approach to Lebesgue integration that defines the integral through the formula $\int f\mathrm d\mu=\lim_n\int f_n\mathrm d\mu$.

What I'm trying to do is to "discover" the definition of "integrable", as if I didn't already know it, by following some suggestions in Lang's book Real and functional analysis. He says that the integration functional ($f\mapsto\int f\mathrm d\mu$) is simply the extension by continuity of the integration functional on the set of integrable simple functions $\mathcal S$, to a larger seminormed space $\mathcal I$, such that $\mathcal S$ is dense in $\mathcal I$. (The notations $\mathcal S$ and $\mathcal I$ are mine. I use $\mathcal M$ for the set of measurable functions). I have interpreted his comments in the following way:

We need to find a vector space $\mathcal I$ and a linear functional $\phi:\mathcal I\to\mathbb C$ such that

1. $\mathcal S\subseteq\mathcal I\subseteq\mathcal M$.
2. $\phi|_\mathcal S$ is the integration functional $f\mapsto\int f\mathrm d\mu$ on $\mathcal S$.
3. $\mathcal I$ is closed under the operation $f\mapsto|f|$.
4. The map $f\mapsto\phi(|f|)$ with domain $\mathcal I$ is a seminorm.
5. $\mathcal S$ is dense in $\mathcal I$, with respect to the seminorm.
6. $\phi$ is continuous with respect to the seminorm.

I use the notation $\rho$ for this seminorm, and $\rho_S$ for the ($L^1$) seminorm already defined on $\mathcal S$.

I want to start with nothing more than this list (or an improved version of it) and the assumption that f is an element of a vector space $\mathcal I$ that satisfies these conditions, and then prove that f satisfies one of the two equivalent definitions of "integrable":

Definition 1: A measurable function f is said to be integrable if there's a sequence $\langle f_n\rangle$ in $\mathcal S$ such that
(a) $\langle f_n\rangle$ is Cauchy with respect to $\rho_S$.
(b) $f_n\to f$ a.e.

Definition 2: A measurable function f is said to be integrable if there's a sequence $\langle f_n\rangle$ in $\mathcal S$ such that
(a) $\langle f_n\rangle$ is Cauchy with respect to $\rho_S$.
(b) $f_n\to f$ in measure.

My attempt goes like this (details available on request): For each n, let $f_n\in\mathcal S$ be such that $\rho(f_n-f)<\frac 1 n$. We have $f_n\to f$ with respect to $\rho$. This implies that $\langle f_n\rangle$ is Cauchy with respect to $\rho_S$. This implies that $\langle f_n\rangle$ is Cauchy in measure. This implies that there's a subsequence $\langle f_{n_k}\rangle$ and a $g\in\mathcal M$ such that $f_{n_k}\to g$ almost uniformly, and therefore in measure and almost everywhere. Since $f_n\to f$ with respect to $\rho$, we also have $f_{n_k}\to f$ with respect to $\rho$, and $\langle f_{n_k}\rangle$ is Cauchy with respect to $\rho_S$.

Now if we can just prove that f=g a.e., then f will satisfy both definitions of "integrable". But I don't know if it can be done without adding more assumptions. It would be trivial if we could use that $L^1$ convergence implies in measure convergence, but I don't see how to prove that without Chebyshev (which relies on the definition that I'm only trying to motivate at this point), or some pretty strong additional assumptions.

 

[micromass]

You are basically trying the Daniell integration approach: http://en.wikipedia.org/wiki/Daniell_integral

Let's try to mimick the proof of Chebyshev.

So from $4$ follows that if $f\geq 0$ and is an element of $\mathcal{I}$ (and thus $f$ is real-valued), then $\varphi(|f|) = \varphi(f)\geq 0$ by property of the seminorm. So we can prove this way that for $f,g\in\mathcal{I}$:

$$f\leq g~\Rightarrow~\varphi(f)\leq \varphi(g)$$

Consider the set $A=\{x~\vert~|f_n(x) - f(x)|\geq \varepsilon\}$. Then $\chi_A$ is in $\mathcal{S}$. It follows from the definition of the integration functional that

$$\mu(A) = \int I_A(x)dx$$

Now it is easy to check that $I_A\leq |f_n - f|$ and both are elements of $\mathcal{I}$. Thus by above follows that

$$\mu(A) = \int I_A(x)dx \leq \varphi(|f_n-f|)$$

Thus follows that if $f_n\rightarrow f$ in $\rho$, then $f_n\rightarrow f$ in measure.

 

[Fredrik]

I don't know how you do it. You're spoiling me, man. Always a perfect answer within an hour, no matter what time of day or night I post. If it ever takes you two hours, I guess I will have figured out which hour of the day that you usually sleep.

 

[micromass]

I don't know how you do it. You're spoiling me, man. Always a perfect answer within an hour, no matter what time of day or night I post. If it ever takes you two hours, I guess I will have figured out which hour of the day that you usually sleep.

Thanks a lot for the kind words! It's just that your threads are always on the mathematics that I really like, so I can't help thinking about it :tongue:

 

[Fredrik]

I found an issue with your proof as I was typing it up for my notes, and I haven't been able to completely resolve it yet. This is what we want to do:
\begin{align}
& E_n=\big\{x\in X:|f_n(x)-f(x)|\geq\varepsilon\big\}\\
& \rho(f_n-f)<\varepsilon\varepsilon '\\
& \mu(E_n)=\int\chi_{E_n}\mathrm d\mu=\phi(\chi_{E_n}) \leq\phi\left(\frac{1}{\varepsilon}|f_n-f|\right) =\frac{1}{\varepsilon}\phi(|f_n-f|) \leq\frac{1}{\varepsilon}\rho(f_n-f)<\varepsilon '.
\end{align} The problem is that the first step of the last calculation is only valid if we already know that $E_n$ is measurable and has finite measure.

I think I've solved the measurability issue. Define $g_n=|f_n-f|$. We know that $g_n\in \mathcal I\subseteq \mathcal M$, because $\mathcal I$ is assumed to be a subset of $\mathcal M$ (functions from $X$ into $\mathbb C$ that are measurable with respect to the Borel algebra on $\mathbb C$) that's closed under the vector space operations and the map $f\mapsto|f|$. Define $F_n=\{z\in\mathbb C|\operatorname{Re} z\geq\varepsilon\}$. Since $F_n$ is closed and $g_n\in\mathcal M$, we have $g_n{}^{-1}(F_n)\in\Sigma$. We also have
$$g_n{}^{-1}(F_n) =\{x\in X|g_n(x)\in F_n\} =\{x\in X|\operatorname{Re}g_n(x)\geq \varepsilon\}=\{x\in X:|f_n(x)-f(x)|\geq\varepsilon\}=E_n.$$
These are my thoughts on the matter of $E_n$ having a finite measure: $E_n$ is the union of the two disjoint sets $\big\{x\in X:f_n(x)\neq 0,\, |f_n(x)-f(x)|\geq\varepsilon\big\}$ and $\big\{x\in X:f_n(x)=0,\, |f_n(x)-f(x)|\geq\varepsilon\big\}$. The former is a subset of $\{x\in X:f_n(x)\neq 0\}$, which has finite measure. The latter is equal to $\{x\in X:|f(x)|\geq\varepsilon\}$. I will denote this set by G. G is measurable, because it's equal to $|f|^{-1}\big(\{z\in\mathbb C:\operatorname{Re} z\geq\varepsilon\}\big)$. I need to prove that G has finite measure.

Suppose that it doesn't. I can derive a false statement from this, if I make assumptions about G, or rather about X. For example: Let M>0 be arbitrary. Let H be an arbitrary measurable subset of G such that $\frac M \varepsilon\leq\mu(H)<\infty$. (Does such a set exist for all M?) We have $|f|\geq\varepsilon\chi_{G}$, and therefore
$$\rho(f)=\phi(|f|) \geq\phi(\varepsilon\chi_{H}) =\varepsilon\phi(\chi_H)=\varepsilon\int\chi_H\mathrm d\mu =\varepsilon\mu(H)\geq M.$$ Since M is arbitrary and $\rho$ is a seminorm on $\mathcal I$, this contradicts the assumption that $f\in\mathcal I$.

Another assumption that yields the desired result is that there's an L>0 such that G is a countable union of disjoint sets that all have a measure that's greater than or equal to L.

 

[micromass]

If $X$ is $\sigma$-finite then this is obvious. Indeed, if $E_n =\bigcup_k E_{n,k}$ is a countable union of finite sets, then you can easily prove like you did that $\mu(E_{k,n})\leq \rho(|f_n-f|)<+\infty$. The fact that $E_n$ has finite measure then follows by taking unions.

The non $\sigma$-finite case is a lot trickier. Define for $A\subseteq X$

$$V(A) = \textrm{sup}\{\phi(g)~\vert~g\in \mathcal{S},~g\leq \chi_A\}$$

You can easily prove that if $A$ has finite measure, then $V(A) = \mu(A)$. It follows that $V(A) = \mu(A)$ for any measurable set $A$.

Now consider the set $G = \{x~\vert~f(x)>a\}$. If $g\in \mathcal{S}$ with $g\leq \chi_G$, then
$$g(x)\leq \frac{1}{a}f(x)$$
It follows that

$$\phi(g)\leq \frac{1}{a}\phi(f)$$

Thus by taking suprema over $g$, it follows that

$$\mu(A) = V(A) \leq \frac{1}{a}\phi(f)$$

I think this does it. Anyway, I would highly recommend you to take a look at "functional analysis" by Lax. He does something very similar to what you want in his first appendix at the end of the book.

 

[Fredrik]

Excellent. I understand the argument. That completes the proof.

I will definitely read that appendix.

 

[micromass]

Excellent. I understand the argument. That completes the proof.

You might want to flesh out the details though. Especially the $\mu(A) = V(A)$. I'm pretty sure it's correct and that I proved it, but it wouldn't be my first mistake, even in this thread :tongue:

 

[Fredrik]

On second thought, it looks like this isn't over yet. At the end of the argument, we have to be able to conclude that since V(G) is finite, $\mu(G)$ is finite. This relies on the earlier claim "For all $A\in\Sigma$ such that $\mu(A)=\infty$, we have $V(A)=\infty$". But can we really prove that, without making assumptions about what sort of subsets A has?

I'm starting to think that we have to assume at least this: For all M>0, every set of infinite measure has a subset of finite measure greater than M.

Don't all interesting measure spaces satisfy this condition anyway?

 

[Fredrik]

I think I don't mind this extra assumption so much. It's a pretty mild one. And this is what I'm going to end up with:

A pair $(\mathcal I,\phi)$ such that $\mathcal I$ is a seminormed space over $\mathbb C$, and $\phi$ is a linear functional on $\mathcal I$, is said to be a fnurgle if it satisfies the conditions on my list. (I just need a term for it until the end of this post). The theorems I'm proving right now will show that for each measure space $(X,\Sigma,\mu)$ that satisfies the additional assumption, there's at most one fnurgle. The theorems also show exactly what set $\mathcal I$ is and what function $\phi$ is.

Then I can show that regardless of whether the measure space satisfies the additional condition, that specific pair $(\mathcal I,\phi)$ is a fnurgle.

So the only problem with the additional assumption is that there may exist some freaky measure space that admits more than one fnurgle.

 

[micromass]

On second thought, it looks like this isn't over yet. At the end of the argument, we have to be able to conclude that since V(G) is finite, $\mu(G)$ is finite. This relies on the earlier claim "For all $A\in\Sigma$ such that $\mu(A)=\infty$, we have $V(A)=\infty$". But can we really prove that, without making assumptions about what sort of subsets A has?

I'm starting to think that we have to assume at least this: For all M>0, every set of infinite measure has a subset of finite measure greater than M.

Don't all interesting measure spaces satisfy this condition anyway?

Yes, for sure.

Alright, here is a counterexample:

Consider $X=\mathbb{R}$, consider all sets measurable and let $\mu(A) = +\infty$ for each nonempty $A\subseteq X$ and of course $\mu(\emptyset) = 0$.

Clearly, $\mathcal{S}= \{0\}$.
Consider the function $f = \chi_\mathbb{R}$. Let $\mathcal{I}=\{\alpha f~\vert~\alpha\in \mathbb{R}\}$. Define $\phi(g) = 0$ for each $g\in \mathcal{I}$. This satisfies all the conditions, but Chebyshev is not satisfied. The function is also not integrable in your sense.

We can eliminate this case either by demanding
(1) There is no measurable subset $E$ of $X$ such that each subset of $E$ either has infinite measure or measure $0$ and such that the measure $E$ is infinite.

(2) Add to the axioms of $\mathcal{I}$ and $\phi$ that if $g_n\in \mathcal{S}$ with $g_n = 0$ a.e., then the limit (if it exists) of the $g_n$ under $rho$ is zero a.e.

(3) Add to the axioms that if $\rho(f)=0$, then $f=0$ a.e. This is similar to (2).

Let's prove that either of those conditions is enough to prove Chebyshev, and thus your theorem.

So take $h$ integrable and $A = \{x~\vert~h(x)>a\}$. It is easy to see that

$$V(A) = \textrm{sup}\{\mu(F)~\vert~F\subseteq A,~\mu(F)<+\infty\}$$

So assume that $V(A)$ is finite but that $\mu(A)$ is infinity, then there exists a sequence $F_n\subseteq A$ such that $\mu(F_n)<+\infty$ and such that $\mu(F_n)\rightarrow V(A)$. We can easily make the sequence increasing, thus we have that $\mu(\bigcup_n F_n) = V(A)$. So there is a set $F$ of measure $V(A)$.

Let's look at $A\setminus F$. Then this has infinite measure. Assume that it contains a set of finite measure $B$. Then $\mu(F\cup B) = \mu(F) + \mu(B)\geq V(A)$. Thus follows that $B$ has measure $0$. So all subsets of $A\setminus F$ either have measure $0$ or measure infinity. If we accepted $(1)$ as true, then $\mu(A\setminus F) = 0$, and thus $\mu(A) = \mu(F)$ has finite measure. Contradiction.

In the other case that $(2)$ or $(3)$ is true, we have found a subset $E\subseteq A$ of infinite measure such that each subset has either measure $0$ or measure infinity.

Consider $\mathcal{I}^\prime = \{f\chi_E~\vert~f\in \mathcal{I}\}$. This can be seen as a quotient space of the vector space $\mathcal{I}$ under the map $P:\mathcal{I}\rightarrow \mathcal{I}^\prime:f\rightarrow f\chi_E$. In particular, it has a canonical seminorm

$$\rho^\prime(f^\prime) = \textrm{inf}\{\rho(f)\in \mathcal{I}~\vert~f\chi_E = f^\prime\}$$

Since $\mathcal{S}$ is dense in $\mathcal{I}$. It follows that $\{g\chi_E~\vert~g\in \mathcal{S}\}$ is dense in $\mathcal{I}^\prime$. But clearly, any integrable function is almost everywhere $0$ on $E$. Thus if $g\in \mathcal{S}$, then $g\chi_E = 0$ a.e. Thus $\rho(g) = 0$ for each $g\in \mathcal{I}^\prime$.

But since $A = \{x~\vert~h(x)>a\}$. We have that $h\chi_E>a$. Thus there exists a sequence $(g_n)_n$ in $\mathcal{S}^\prime$ such that $g_n\rightarrow h\chi_E$ in $\rho^\prime$. Since $g_n=0$ a.e., it follows from $(2)$that $h\chi_E = 0$ a.e, which contradicts $h\chi_E>a$.

If $(3)$ is true, then it follows from $\rho(g)=0$ for each $g\in \mathcal{I}^\prime$ that $h\chi_E = 0$ a.e. which contradicts again $h\chi_E>a$.

 

[Fredrik]

Thanks again. This is very helpful. I agree that you have found a counterexample. I will have to think about which additional assumption I like best, but I think we have solved the problem now. It would have taken me at least another week to figure all of this out (and I might not have figured it out at all), so I really appreciate your contribution.

 

[Fredrik]

$$V(A) = \textrm{sup}\{\mu(F)~\vert~F\subseteq A,~\mu(F)<+\infty\}$$

So assume that $V(A)$ is finite but that $\mu(A)$ is infinity, then there exists a sequence $F_n\subseteq A$ such that $\mu(F_n)<+\infty$ and such that $\mu(F_n)\rightarrow V(A)$. We can easily make the sequence increasing, thus we have that $\mu(\bigcup_n F_n) = V(A)$. So there is a set $F$ of measure $V(A)$.

I'd like to understand this part. I'm with you until you say that $\mu\left(\bigcup_n F_n\right)=V(A)$. I see how to construct an increasing sequence $\langle F_n\rangle_{n=1}^\infty$ in $\Sigma$ such that $\mu(F_n)\to V(A)$, but this doesn't seem to imply that $\mu\left(\bigcup_n F_n\right)=V(A)$. If $\bigcup_n F_n$ has finite measure, then we can show something like $V(A)-\frac 1 n<\mu\left(\bigcup_n F_n\right)\leq V(A)$. But what if $\bigcup_n F_n$ has infinite measure?

If the sequence had been mutually disjoint (mine isn't), then we could have said that the equalities $\infty=\mu\left(\bigcup_n F_n\right) =\sum_n\mu(F_n)$ imply that some partial sum is greater than V(A), and this would be a falsehood that proves that the first equality is actually wrong.

What I actually want to do is to show that your description of what I'll call a "non-ridiculous" measure space

There's no set A such that $\mu(A)=\infty$ and all subsets of A have measure 0 or ∞.​

implies my version of it

For all A such that $\mu(A)=\infty$, and all M>0, there's a $B\subseteq A$ such that $M\leq\mu(B)<\infty$.​

I figured it would be a good strategy to show that the set $\big\{\mu(F)|F\subseteq E,\, \mu(F)<\infty\big\}$ isn't bounded from above, by attempting to derive a contradiction from the assumption that it is.

I also tried a different strategy, but got stuck on a similar detail: For each $n\in\mathbb Z^+$, let $F_n$ be a subset of $E-\bigcup_{k=1}^n F_k$ with finite measure. Define $F=\bigcup_{k=1}^\infty F_k$. If $\mu(F)=\infty$ there's an n such that $M\leq\mu\left(\bigcup_{k=1}^n F_k\right)<\infty$. But if $\mu(F)<\infty$, I don't see a way to proceed.

 

[micromass]

I'd like to understand this part. I'm with you until you say that $\mu\left(\bigcup_n F_n\right)=V(A)$. I see how to construct an increasing sequence $\langle F_n\rangle_{n=1}^\infty$ in $\Sigma$ such that $\mu(F_n)\to V(A)$, but this doesn't seem to imply that $\mu\left(\bigcup_n F_n\right)=V(A)$. If $\bigcup_n F_n$ has finite measure, then we can show something like $V(A)-\frac 1 n<\mu\left(\bigcup_n F_n\right)\leq V(A)$. But what if $\bigcup_n F_n$ has infinite measure?

I am applying Theorem 1.2.1.(iii) in Friedman. It says immediately that

$$\mu(F_n)\rightarrow \mu\left(\bigcup_n F_n\right)$$

and since $\mu(F_n)\rightarrow V(A)$, we have equality.

What I actually want to do is to show that your description of what I'll call a "non-ridiculous" measure space

There's no set A such that $\mu(A)=\infty$ and all subsets of A have measure 0 or ∞.​

implies my version of it

For all A such that $\mu(A)=\infty$, and all M>0, there's a $B\subseteq A$ such that $M\leq\mu(B)<\infty$.​

I figured it would be a good strategy to show that the set $\big\{\mu(F)|F\subseteq E,\, \mu(F)<\infty\big\}$ isn't bounded from above, by attempting to derive a contradiction from the assumption that it is.

That should work. A proof like I did in my last post should do it, no? Well, that is if you didn't find some fatal flaw in it, which is entirely possible.

 

[Fredrik]

Thanks. I'll study the proof of that theorem right away. The rest of the proof from your previous post looks fine to me. My only concern was that F might have infinite measure.

Edit: OK, the proof of the step I was stuck on is essentially just the standard trick to rewrite a countable union as a countable union of mutually disjoint sets, and then use that $\mu(F_n)\to V(A)$.