Instead of $\mu (X)<\infty$ suppose that $|f_n|\leqslant g, \forall n\in\mathbb N,$ and $g\in L^1(μ) $. $$$$ If $f_n\longrightarrow f $ a.e. in X , then prove that:$$\forall \epsilon>0, \exists E\subset X,s.t. \mu (E)<\epsilon$$ and $$f_n\longrightarrow f $$uniformly on $E^c$.

A hint which accompanies this exercise and I want to use is : Use the sets $\left\{ g>1\right\} $ and $\left\{ 2^{-k}<g\leqslant2^{1-k}\right\}$ which have finite measures due to the integrability of g.

I am thinking of applying Egoroff's theorem to each of these sets. We can write:

$$X=\{ g>1\}\cup(\bigcup\limits_{k=1}^{\infty}\left\{ 2^{-k}<g\leqslant2^{1-k}\right\})\cup \{g=0\}$$

Then there exist $A\subset G=\{g>1\}$ ,and $A_k\subset G_k=\left\{ 2^{-k}<g\leqslant2^{1-k}\right\}, $ such that: $$\mu (A)<\varepsilon , \mu (A_k)<\varepsilon $$ and $$f_n\longrightarrow f$$uniformly in $E^c=(G\setminus A)\cup(\bigcup\limits_{k=1}^{\infty}G_k\setminus A_k)\cup \{f_n=0,\forall n\}$, where the last set in the union is a superset of $\{g=0\}$.

However, I fail to prove that $\mu (E)<\varepsilon$. Maybe it is a simple calculation, or a better choice of sufficiently small $\varepsilon$'s . I would appreciate your help.

  • 1
    $\begingroup$ If you picked $\mu(A) < \frac{\epsilon}{2}, \mu(A_k) < \frac{\epsilon}{2^{k+1}}$ Then $\mu(E) \le \mu(A\cup \bigcup A_k) \le \epsilon$. $\endgroup$ – spenceryue May 15 '18 at 19:42
  • $\begingroup$ But I think $f_n$ would be converging at entirely different rates across all your sets $A_k$ so that no uniform $N$ can be found so that $|f_n(x) - f(x)| < \eta$ for all $x\in E^c$ and $n>N$. $\endgroup$ – spenceryue May 15 '18 at 21:22

Notice that $E^c\supset G\setminus A$, hence $E\subset A$. We can conclude, since $\mu$ is a positive measure and $\mu(A)<\varepsilon$.

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