3
$\begingroup$

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.

$\endgroup$
  • 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
1
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.