Closures of the set of measurable functions

[Fredrik]

Can a measurable function be a.e. equal to a non-measurable function?

Let $(X,\Sigma,\mu)$ be an arbitrary measure space. Let M be the set of measurable functions from X into $\mathbb C$. I know that M is closed under pointwise limits. I'd like to know if M is also closed under the types of limit that involve a measure: "almost everywhere", "in measure" and "almost uniformly".

I think I have proved that if $f$ is a function from X into $\mathbb C$, and $\langle f_n\rangle_{n=1}^\infty$ is a sequence in M such that $f_n\to f$ a.e., then there's a $g\in M$ such that f=g a.e. (I think I have also proved similar results for the other types of limit). If the two statements $g\in M$ and f=g a.e. together imply that $f\in M$, this would mean that M is closed under a.e. limits. Is it?

 

[disregardthat]

The (unmeasurable) indicator function on an unmeasurable null set is equal a.e. to the constant function 0 on all of X. I'm not entirely sure about the conventions regarding whether null sets are considered measurable by default or not, but this seems to me like a counter-example to your suggestion.

 

[micromass]

The (unmeasurable) indicator function on an unmeasurable null set is equal a.e. to the constant function 0 on all of X. I'm not entirely sure about the conventions regarding whether null sets are considered measurable by default or not, but this seems to me like a counter-example to your suggestion.

Yes, but that's because the Borel sets are incomplete under Lebesgue measure (or the specific measure under consideration). If we use the Lebesgue measurable sets, then the problem does not occur.

 

[Fredrik]

The (unmeasurable) indicator function on an unmeasurable null set is equal a.e. to the constant function 0 on all of X. I'm not entirely sure about the conventions regarding whether null sets are considered measurable by default or not, but this seems to me like a counter-example to your suggestion.

A measure is said to be complete if every subset of a set of measure zero has measure zero. The Lebesgue measure is complete, but I'd like to stay as general as possible and not assume that my $\mu$ is complete.

This also means that my definition of f=g a.e. isn't that $\mu\big(\{x\in X:f(x)\neq g(x)\}\big)=0$. It's that $\{x\in X:f(x)\neq g(x)\}$ is a subset of a set of measure zero, or equivalently, that there exists a set E such that $\mu(E)=0$ and $f(x)=g(x)$ for all $x\in E^c$.

OK, I see that if S is an unmeasurable subset of a set of measure zero, then $\chi_S=0$ a.e. but $\chi_S$ is not measurable, because the preimage of an open ball that contains 1 is S. Good example, assuming that sets like S exist. (I don't know any examples of such sets, but if they exist, I can probably find them in any book on measure and integration theory). I guess that settles it when $\mu$ is such that sets like S exist.

But sets like S can only exist if $\mu$ is not complete. So I still don't know how to handle the general case.

 

[disregardthat]

Does the following argument work for the complete case:

Let $f = g$ on the complement $N^c$ of a null set $N$. Then for any measurable set $E$ of $\mathbb{C}$, $$f^{-1}(E) = (f|_{N^c})^{-1}(E) \cup (f|_N)^{-1}(E) = (g|_{N^c})^{-1}(E) \cup P,$$ where $(g|_{N^c})^{-1}(E) = g^{-1}(E) \cap N^c$ and $P \subseteq N$ is a null-set.

If $\mu$ is a complete measure, then $P$ is measurable, and $g^{-1}(E) \cap N^c$ is measurable as well, since its complement is $g^{-1}(E)^c \cup N$. Hence $f$ is a measurable function.

 

[disregardthat]

It's been some time since I've been working with measure theory. A subset of X is unmeasurable with respect to a measure on a sigma algebra on X if it's not an element of the sigma algebra, right? If so, consider the following argument:

Take any $\sigma$-algebra $\Sigma$ on a set $X$ such that there exists unmeasurable subsets of $X$. Define a measure by $\mu(E) = 0$ for all $E \in \Sigma$. Then every unmeasurable set is a subset of the null set $X$.

 

[Fredrik]

Does the following argument work for the complete case:

Let $f = g$ on the complement $N^c$ of a null set $N$. Then for any measurable set $E$ of $\mathbb{C}$, $$f^{-1}(E) = (f|_{N^c})^{-1}(E) \cup (f|_N)^{-1}(E) = (g|_{N^c})^{-1}(E) \cup P,$$ where $(g|_{N^c})^{-1}(E) = g^{-1}(E) \cap N^c$ and $P \subseteq N$ is a null-set.

If $\mu$ is a complete measure, then $P$ is measurable, and $g^{-1}(E) \cap N^c$ is measurable as well, since its complement is $g^{-1}(E)^c \cup N$. Hence $f$ is a measurable function.

Yes, that looks good, except for the minor detail that $(g|_{N^c})^{-1}(E)$ isn't equal to $g^{-1}(E)\cap N^c$. What's important is that the former set is of the form $g^{-1}(E)\cap Q^c$, where $Q$ is a null set. Since a $\sigma$-algebra is closed under unions, intersections and complements, this is sufficient to ensure that $f^{-1}(E)\in\Sigma$.

It's been some time since I've been working with measure theory. A subset of X is unmeasurable with respect to a measure on a sigma algebra on X if it's not an element of the sigma algebra, right? If so, consider the following argument:

Take any $\sigma$-algebra $\Sigma$ on a set $X$ such that there exists unmeasurable subsets of $X$. Define a measure by $\mu(E) = 0$ for all $E \in \Sigma$. Then every unmeasurable set is a subset of the null set $X$.

A measure that takes everything to 0? I don't think that's useful. The set of functions that are measurable with respect to that measure is different from the set of functions that are measurable with respect to my arbitrary measure.

There is however a standard procedure to define a complete measure from an incomplete one. It's described here. (Page 12, beginning of section 1.5). If the link doesn't work for you, try replacing .com with your country domain name.

 

[disregardthat]

My point was only that in my example there exists unmeasurable null-sets, to conclude that f = 0 a.e. does not imply that f is measurable in a general setting.

EDIT: Okay, I assumed you were asking for whether $g \in M$ and f = g a.e. $\Rightarrow f \in M$. But I'm not sure if that's what you want.

 

[disregardthat]

Yes, that looks good, except for the minor detail that $(g|_{N^c})^{-1}(E)$ isn't equal to $g^{-1}(E)\cap N^c$. What's important is that the former set is of the form $g^{-1}(E)\cap Q^c$, where $Q$ is a null set. Since a $\sigma$-algebra is closed under unions, intersections and complements, this is sufficient to ensure that $f^{-1}(E)\in\Sigma$.

Maybe it's just me, but I don't see how $(g|_{N^c})^{-1}(E)$ is not equal to $g^{-1}(E)\cap N^c$. Could you explain that a bit further?

 

[Fredrik]

My point was only that in my example there exists unmeasurable null-sets, to conclude that f = 0 a.e. does not imply that f is measurable in a general setting.

Ah, that's a good point.

EDIT: Okay, I assumed you were asking for whether $g \in M$ and f = g a.e. $\Rightarrow f \in M$. But I'm not sure if that's what you want.

That is the main thing that I was concerned about. I think I got a little confused along the way. I see now that you have proved that this implication doesn't hold in general.

Maybe it's just me, but I don't see how $(g|_{N^c})^{-1}(E)$ is not equal to $g^{-1}(E)\cap N^c$. Could you explain that a bit further?

OK, I agree. I first wrote down $(g|_{N^c})^{-1}(E)=g^{-1}(E)-\{x\in N|g(x)\in E\}$, which looked different from $g^{-1}(E)-N$. But I see now that there's no difference.

 

[Fredrik]

I think I understand the other issue too, the issue of whether M is closed under any of these types of limits. I think the answer is that if a sequence in M is convergent in any of the three ways (almost everywhere, in measure, almost uniformly), then the sequence has multiple limits, and we can be sure that at least one of them is in M, but in the general case, we can't be sure that all of them are. The other limits will however always be a.e. equal to a measurable function.

Thank you very much. Your posts were very helpful.

 

[disregardthat]

Back to your original intention, as I understand it you want to take limits of sequences of equivalence classes of measurable functions $\{[f_n]\}$, where the measurable functions f ~ g if f = g a.e.

A sequence $\{[f_n]\}$ has a well-defined limit [f], not necessarily measurable (note: the equivalence classes [f] and $[f_n]$ are with respect to different equivalence relations. The first is an equivalence class of a.e. equal measurable functions, the other an equivalence class of a.e. equal functions in general): If $\lim_{n \to \infty} f_n(x)$ exist a.e., say on $N^c$ where $N$ is a null-set, then for any other sequence of representatives $\{g_n\}$ of the sequence, $\lim_{n \to \infty} g_n(x)$ also exists a.e (because: if $g_n$ and $f_n$ agree on $(N_n)^c$ where $N_n$ is a null-set, then $\lim_{n \to \infty} g_n(x)$ exists and is equal to $\lim_{n \to \infty} f_n(x)$ for $x$ in $(N \cup \bigcup_n N_n)^c$ where $N \cup \bigcup_n N_n$ is a null-set. Thus the function f defined by $\lim_{n \to \infty} f_n(x)$ on $N^c$ and $0$ elsewhere is well-defined up to a.e. equivalence.

If you have proved that there exists a measurable function g such that g = f a.e., then [g] is a well-defined limit of the sequence $\{[f_n]\}$ of measurable functions.

So I don't think you need actually the condition that $g \in M$ and f = g a.e. $\Rightarrow f \in M$. How did you prove that such a g exists?

 

[Fredrik]

Back to your original intention, as I understand it you want to take limits of sequences of equivalence classes of measurable functions $\{[f_n]\}$, where the measurable functions f ~ g if f = g a.e.

I've had some thoughts along these lines, but I wouldn't say that it was my original intention. I just wanted to know to what extent M is closed under these types of limits.

So I don't think you need actually the condition that $g \in M$ and f = g a.e. $\Rightarrow f \in M$. How did you prove that such a g exists?

The difficult theorem to prove is that M is closed under pointwise limits. Once you have that result, you can do this:

Theorem: Let $\langle f_n\rangle_{n=1}^\infty$ be an arbitrary sequence in M. Let $f:X\to\mathbb C$ be arbitrary. If $f_n\to f$ a.e, then there's a $g\in M$ such that $f=g$ a.e.

Proof: Let E be a set such that $\mu(E)=0$ and $f_n\to f$ pointwise on $E^c$. Define $g:X\to\mathbb C$ by
$$
g(x)=
\begin{cases}
f(x) & \text { if }x\in E^c\\
0 & \text { if }x\in E.
\end{cases}
$$ This g is clearly a.e. equal to f, and it's measurable because it's the pointwise limit of the sequence $\langle\chi_{E^c}f_n\rangle_{n=1}^\infty$.

Ah yes, you also have to prove that M is closed under products. Now that I think about it, I'm not even sure that it is. I need to think about that.

 

[micromass]

Ah yes, you also have to prove that M is closed under products. Now that I think about it, I'm not even sure that it is. I need to think about that.

No problem there since $\mathbb{C}\times\mathbb{C}\rightarrow \mathbb{C}:(x,y)\rightarrow xy$ is continuous and thus measurable. Now $fg$ can be defined as the composition $x\rightarrow (f(x),g(x))\rightarrow f(x)g(x)$ which is then measurable.