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Exercise 40 - Show in Egoroff's theorem, the hypothesis "$\mu(X)<\infty$" can be replaced by "$|f_n|\le g$ for all $n$, where $g \in L^1(\mu)$."

Egoroff's Theorem - Suppose $\mu(X)<\infty$, and $f_1, f_2, ...$ and $f$ are measurable complex-valued functions on $X$ such that $f_n \rightarrow f$ a.e. Then for every $\epsilon > 0$ there exists $E\subset X$ such that $\mu(E)<\epsilon$ and $f_n \rightarrow f$ uniformly on $E^c$.

The proof of the original Egoroff's theorem uses continuity from above, which requires the finiteness assumption. I tried to use the dominating $g$ to get some finiteness condition.

When $\mu$ is $\sigma$-finite, I showed the integral of $g$ is concentrated on some set of finite measure. But this does not seem to help proving the statement since the complement of this set still have infinite measure (if $X$ has).

Can anyone give me some hint on the problem? Thank you

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  • $\begingroup$ The statement that support of $L^1$ function $g$ is of finite measure is false, but instead it is always a $\sigma$-finite set. $\endgroup$ – Song Nov 7 '18 at 2:50
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Fix $\varepsilon>0$. For each $N \geq 1$, apply the original version restricted to the set $G_N:=\{g \geq 1/N\}$ (which clearly has finite measure if $\int g \, d\mu<\infty$), in order to get a set $E_N \subset G_N$ with $\mu(E_N)<\frac{\varepsilon}{2^N}$ such that $f_n \to f$ uniformly on $G_N \!\setminus\! E_N$. Now let $E:=\bigcup_{N=1}^\infty E_N$.

Obviously $\mu(E)<\varepsilon$. It's not too difficult to show that $f_n \to f$ uniformly on $X \!\setminus\! E$.

[If you're having trouble with this last bit: First, as an exercise, show that $f_n(x)=\mathbf{1}_{[-n,n]}(x)e^{-|x|}$ converges to $f(x)=e^{-|x|}$ uniformly on $\mathbb{R}$. Then generalise the reasoning.]

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