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

Proof - Let $\mu(X) < \infty$ and $\{f_j\}_{j\in\mathbb{N}}$ and $f$ be measurable complex-valued functions on $X$ such that $f_n\rightarrow f$ a.e. Then, since $f_n\rightarrow f$ a,e, there exists a set $E\in M$ such that $\mu(E) = 0$ and for all $x\in E^c$, $f_n(x) \rightarrow f(x)$. So we have $f_n\chi_{E_c}\rightarrow fX_{E^c}$ and $f\chi_{E^c} = f$ a.e.

Now without loss of generality assume $f_n\rightarrow f$ pointwise. For $n,k$ define $E_n(k) = \bigcup_{m=k}^{\infty}\{|f_m - f|\geq 1/k\}$. Then clearly $E_n(k)$ is decreasing since $E_{n+1}(k)\subseteq E_n(k)$ and $\bigcap_{n=1}^{\infty}E_n = \emptyset$ by pointwise converence. So by continuity from above $\lim_{n\rightarrow \infty}\mu(E_n(k)) = 0$.

Let $\epsilon > 0$, for each $k$, choose $n_k$ such that $\mu(E_{n_k}) < \epsilon 2^{-k}$. Set $E = \bigcup_{1}^{\infty}E_{n_k}(k)$ then $\mu(E) < \epsilon$. So we have uniform convergence on $X\setminus E$.

I tried to follow what Folland did but I am not sure if this is correct completely and I honestly don't really like his approach. If there is an alternative way of proving this theorem any suggestions would be greatly appreciated. Also note that the first paragraph is not used, although it is still true.

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

Proof - Let $\mu(X) < \infty$ and $\{f_j\}_{j\in\mathbb{N}}$ and $f$ be measurable complex-valued functions on $X$ such that $f_n\rightarrow f$ a.e.

Now, note that $f_n(x)$ does NOT converge to $f(x)$, if and only if there is $k \in \mathbb{N}$ $k>0$, such that for all $n\in \mathbb{N}$, there is $m\geq n$ such that $|f_n(x)-f(x)|\geq 1/k$.

In other words: $f_n(x)$ does NOT converge to $f(x)$, if and only if there is $k \in \mathbb{N}$ $k>0$, such that $x \in\bigcap_{n=1}^\infty\bigcup_{m=n}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\}$

Since $f_n(x)$ does NOT converge to $f(x)$ except in a set of measure zero, we have that $$\mu \left(\bigcap_{n=1}^\infty\bigcup_{m=n}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\} \right) =0$$

Note that $\{ \bigcup_{m=n}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\} \}$ is a monotone non-increasing sequence of measurable sets, and since $\mu(X)<\infty$, using the continuity of the measure from above, we have

$$\lim_{n \to \infty}\mu \left(\bigcup_{m=n}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\} \right) = \mu \left(\bigcap_{n=1}^\infty\bigcup_{m=n}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\} \right) =0$$

So, given $\epsilon>0$ and $k\in\mathbb{N}$, $k\geq 1$, we can choose $n_k$ large enough to ensure $$\mu \left(\bigcup_{m=n_k}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\} \right) \leq \epsilon / 2^k$$

So it follows that

$$\mu \left(\bigcup_{k=1}^\infty\bigcup_{m=n_k}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\} \right) \leq \sum_{k=1}^\infty\mu \left(\bigcup_{m=n_k}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\} \right) \leq \epsilon $$

Now, note that, if $y\notin \bigcup_{k=1}^\infty\bigcup_{m=n_k}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\}$ then for all $k\geq 1$ and for all $m\geq n_k$, $|f_m(y) -f(y)| < 1/k$. It means that $f_n$ converges to $f$ uniformly on the complement of $\bigcup_{k=1}^\infty\bigcup_{m=n_k}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\}$.

Make $E= \bigcup_{k=1}^\infty\bigcup_{m=n_k}^\infty\{x : |f_m(x)-f(x)|\geq 1/k\}$.

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Your proof appears to be mostly correct, aside from a few typos. I believe your function $E_n(k)$ should be defined by $$E_n(k)=\cup_{m=n}^\infty\{|f_m-f|\geq 1/k\} $$

As far as other proofs are concerned, I haven't seen much variation in them. Here is have one from an exercise in Rudin's Real and Complex Analysis (I believe it is in Chapter 3):

Egoroff's Theorem If $\mu(X)<\infty$, if $\{f_n\}$ is a sequence of complex measurable functions which converges pointwise on $X$, and if $\varepsilon>0$, then there is some measurable $E\subset X$ with $\mu(E-X)<\varepsilon$ such that $\{f_n\}$ converges uniformly on $E$.

With the following proof as suggested in the exercise:

For $k,n\in\mathbb{N}$, put \begin{align*} S(n,k)=\cap_{i,j>n}\{x\in X:|f_i(x)-f_j(x)|<\frac{1}{k}\}. \end{align*} Fix $k\in\mathbb{N}$. For each $x\in X$, there is some $N\in\mathbb{N}$ such that $|f_i(x)-f_j(x)|<1/k$ whenever $i,j>N$, so $x\in S(N,k)$ and thus \begin{align*} X=\cup_{n=1}^\infty S(n,k). \end{align*} Since $S(n,k)\subset S(m,k)$ whenever $m>n$, we have \begin{align*} \mu(X)=\lim_{n\to\infty}\mu(S(n,k)) \end{align*} Thus for each $k$, there is some $N_k\in\mathbb{N}$ such that \begin{align*} \mu(X-S(n,k))<\frac{\varepsilon}{2^k} \end{align*} whenever $n\geq N_k$. Put $E=\cap_{k=1}^\infty S(N_k,k)$. Then we have \begin{align*} \mu(X-E)=\mu\left(\bigcup_{k=1}^\infty(X-S(N_k,k))\right) \leq\sum_{k=1}^\infty\mu(X-S(N_k,k)) <\varepsilon. \end{align*} Clearly $\{f_n\}$ is uniformly Cauchy on $E$, and it is therefore uniformly convergent on $E$.

As you can see, there is a lot of similarity between the two proofs.

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