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Consider a sequence of measurable functions $f_{n}:[0,1]\rightarrow\mathbb{R}$ such that $f_{n}\rightarrow0$ pointwise almost everywhere. Prove that for every $\epsilon>0$ there exists $A\subset[0,1]$ such that $m(A)<\epsilon$, $f_{n}$ are all integrable on $[0,1]\backslash A$, and $\int_{[0,1]\backslash A}f_{n}\rightarrow0$ as $n\rightarrow\infty$.

The statement I'm trying to prove above is really similar to Egoroff's theorem. I know that the Egoroff's theorem states that if the first part is true, then it is uniformly convergent on $[0,1]\backslash A$. How would this be possible?

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Hint: Suppose $f_1,f_2,\dots:[0,1]\to \mathbb R$ are measurable and $\epsilon>0.$ Show there exists a set $B,m(B)<\epsilon,$ such that each $f_n$ is bounded on $[0,1]\setminus B.$

Added later: Here's how to use the hint for the given sequence $f_1,f_2,\dots$ Let $\epsilon>0.$ First we find $B$ as above, with $\epsilon/2$ in place of $\epsilon.$ Second, by Egoroff, we know there is $C$ with $m(C)<\epsilon/2$ such that $f_n\to 0$ uniformly on $[0,1]\setminus C.$

Let $A=B\cup C.$ Then $m(A)<\epsilon.$ On $[0,1]\setminus A,$ each $f_n$ is bounded, and $f_n\to 0$ uniformly. Now on sets of finite measure, such as $[0,1]\setminus A,$ boundedness implies integrability, and uniform convergence implies $L^1$ convergence. Thus we're done, having proved a result stronger than asked for.

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  • $\begingroup$ Not every bounded function is integrable. Can you please elaborate how this can help? $\endgroup$ – Ya G May 8 '18 at 3:12
  • $\begingroup$ Every bounded function is integrable over a set of finite measure. $\endgroup$ – zhw. May 8 '18 at 3:35
  • $\begingroup$ I addded to my answer. $\endgroup$ – zhw. May 8 '18 at 15:52
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On $[0,1]-A$, $(f_{n})$ is uniformly convergent to $0$, so given $\eta>0$, one finds some $N$ such that $|f_{n}(x)|<\eta$ for all $x\in[0,1]-A$ and $n\geq N$, then $\displaystyle\int_{[0,1]-A}|f_{n}(x)|dx\leq\int_{[0,1]-A}\eta dx=\eta m([0,1]-A)\leq\eta\cdot 1<\infty$ for all such $n$, the integrability follows for all $f_{n}$ with $n\geq N$, this also proves that $\displaystyle\int_{[0,1]-A}|f_{n}(x)|dx\rightarrow\infty$ as $n\rightarrow\infty$.

General Case:

Find some $M_{x}>0$ such that $|f_{k}(x)|\leq M_{x}$ for all $k$. Let $\epsilon>0$ be given. Now we write that \begin{align*} [0,1]=\bigcup_{M}S_{M} \end{align*} where \begin{align*} S_{M}=\{x\in[0,1]: |f_{k}(x)|\leq M~\text{for all }k\}, \end{align*} then $m(S_{M})\uparrow 1$ and we have some $M$ such that $m([0,1]-S_{M})<\epsilon$ and hence $|f_{k}(x)|\leq M$ for all such $k$.

Now apply Egorov Theorem to that $[0,1]-S_{M}$.

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  • $\begingroup$ How does this show that all $f_n$ are integrable? Surely it only gives integrability for $n \geq N$? $\endgroup$ – B. Mehta May 8 '18 at 2:01

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