Egoroff's Theorem: Finite Measure Convergence

[Bashyboy]

Homework Statement

Show that Egoroff's theorem continues to hold if the convergence is pointwise a.e. and $f$ is finite a.e.

Homework Equations

Here is the statement of Egoroff's theorem:

Assume that $E$ has finite measure. Let $\{f_n\}$ be a sequence of measurable functions on $E$ that converges pointwise on $E$ to the real-valued function $f$. Then for each $\epsilon > 0$, there is a closed set $F$ contained in $E$ for which $f_n \to f$ uniformly on $F$ and $m(E-F) < \epsilon$.

The Attempt at a Solution

Am I allowed to use Egoroff's theorem to prove that statement?The proof of Egoroff's theorem doesn't presuppose it, so I am wondering if Egoroff's theorem is one of those theorems where the special case can actually be used to prove the general case. In fact, after the proof of Egoroff's theorem, the author writes "It is clear that Egoroff's theorem also holds if the convergence is pointwise a.e. and the limit function is finite a.e." The words "it is clear" usually indicate that the problem is easy; otherwise, this seems like it would be a pretty hard problem.

 

[andrewkirk]

It seems like a rather strange way to state the problem. What is the value of $f$ at domain points where it is not finite? Are we to assume that the range of $f$ is the extended real numbers? What if the set of points where the $f_n$ do not pointwise converge doesn't match the set of points where $f$ is 'not finite'?

In short, I am not convinced that the proposition you are being asked to prove is well-defined.

Here's an alternative, that is well-defined, and that may be what they want you to prove:

Assume that $E$ has finite measure and has measurable subset $U$ with $m(U)=m(E)$. Let $\{f_n\}$ be a sequence of measurable functions on $E$, and $f$ be a real-valued function on $U$ such that $\{f_n|_U\}$ converges pointwise to $f$. Then for each $\epsilon > 0$, there is a closed set $F\subseteq U$ for which $f_n \to f$ uniformly on $F$ and $m(U-F) < \epsilon$.​

Do you think that's what they mean?

It would be perfectly valid to use Egoroff's theorem to prove this extension, as long as the functions to which Egoroff's theorem was applied (a) differed from those for which we are trying to prove the extension and (b) satisfied the premises of the base Egoroff theorem.

Alternatively, it may be that an inspection of the author's proof of Egoroff's theorem shows that we did not use the full strength of its premises, and that weaker premises of the type I wrote above would allow the original proof to survive without adjustment.

 

[Bashyboy]

It seems like a rather strange way to state the problem. What is the value of fff at domain points where it is not finite? Are we to assume that the range of fff is the extended real numbers?

Yes. At those values, $f$ will take on an extended real value (i.e., $\pm \infty$).

What if the set of points where the fnfnf_n do not pointwise converge doesn't match the set of points where fff is 'not finite'?

Then just subtract them, which I will explain below.

Do you think that's what they mean?

Hmmm...It's possible. I took the problem statement mean that I am to prove the following: Assume that $m(E)< \infty$. Let $f_n : E \to \Bbb{R}$ converge pointwise to $f$ on $E-A$, where $A \subseteq E$ has zero measure, and where $f$ is finite over $E-B$ with $B \subseteq E$ of zero measure.

And here is what I had in mind for proving it: Note that $f$ is finite over $E-(A \cup B)$ and $f_n$ converges to $f$ over $E-(A \cup B)$, so that we can apply Egoroff's theorem. Also, $m(A \cup B) \le m(A) + m(B) = 0$, so that $m(E-(A \cup B)) = m(E)$. Since $A \cup B \subseteq E$ is of zero measure, it is a measurable set and therefore the $f_n$ restricted to $E-(A \cup B)$ is measurable, which means that $f$ is measurable on $E-(A \cup B)$, too. Given $\epsilon > 0$, Egoroff's theorem says there exists a closed set $F$ such that $F \subseteq E - (A \cup B) \subseteq E$ such that $f_n \to f$ uniformly on $F$ and $m((E-(A \cup B) - F) < \epsilon$. The set $F$ being closed means it is measurable, and has finite outer measure, so that $m(E-F) = m(E)-m(F) = m(E-(A \cup B)) - m(F)) = m((E-(A \cup B)) - F) < \epsilon$, where the excision property was used a few times. Hence $m(E-F) < \epsilon$ and $f_n \to f$ uniformly on $F$.

Does that seem right?

 

[andrewkirk]

It looks good. The only gap I can see that needs to be filled is that our application of Egoroff tells us that $F$ is closed in $E-(A\cup B)$. We need to show that that means it is also closed in $E$.