# Real Analysis, Folland Excercise 2.40

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

• The statement that support of $L^1$ function $g$ is of finite measure is false, but instead it is always a $\sigma$-finite set. – Song Nov 7 '18 at 2:50

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.]