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user193319
user193319
13:00
Let $A$ be a Lebesgue measurable set of finite measure, and let $\{f_n : A \to \Bbb{R}\}$ be a sequence of measurable sets converging to $f : A \to \Bbb{R}$ pointwise. Does there exist $B \subseteq A$ such that $m(A \setminus B) = 0$ and $f_n$ converges uniformly to $f$ on $B$? I think I was able to prove this using Egoroff's theorem. I just want to verify that it is in fact true.
Martin Sleziak
Martin Sleziak
13:23
@user193319 Isn't $f_n(x)=x^n$ defined on $[0,1]$ a counterexample to this claim?
user193319
user193319
Hmm...I think so.
Martin Sleziak
Martin Sleziak
I see that I was too slow:
in Mathematics, 5 mins ago, by MatheinBoulomenos
@user193319 wait, no, I made a mistake, this is not true. Consider $A=[0,1)$ and $f_n=x^n$, this converges pointwise to the zero function.
But if $B \subset [0,1)$ is of measure zero, then for any $m \in \Bbb N$, there is a point in the interval $[1-1/m,1-1/(m+1)]$ which is not in $B$ (because else $B$ would have positive measure). Using that each $f_n$ is monotonic, we get that $\sup_{x \in [0,1) \setminus B} |f_n(x)| \geq (1-1/m)^n$, letting $m \to \infty$ gives $\sup_{x \in [0,1) \setminus B} |f_n(x)| \geq 1$ which implies that $f_n$ doesn't converge uniformly to $0$ on $[0,1) \setminus B$
in Mathematics, 1 min ago, by MatheinBoulomenos
corrected version: if $B \subset [0,1)$ is a subset such that $[0,1) \setminus B$ is of measure $0$, then for any $m \in \Bbb N$, there is a point in the interval $[1-1/m,1-1/(m+1)]$ which is in $B$ (because else $[0,1) \setminus B$ would have positive measure). Using that each $f_n$ is monotonic, we get that $\sup_{x \in B} |f_n(x)| \geq (1-1/m)^n$, letting $m \to \infty$ gives $\sup_{x \in B} |f_n(x)| \geq 1$ which implies that $f_n$ doesn't converge uniformly to $0$ on $B$
I think that basically the main point here is that if you omit any null-set from $[0,1]$, you will still have some points that are arbitrarily close to $1$. Which are the points that cause problems with uniform convergence.

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