0
$\begingroup$

Another old exam problem in measure theory im not sure about. Let $(X,A,\mu)$ be a measure space and $f,g, f_n, g_n$ measurable functions on $X$ such that: $(f_n)$ converges to $f$ in $L^1(\mu)$ and $g_n \leq 2$ and $g_n$ converges to g $\mu -a.e$

Show that $(f_ng_n)$ converges to $fg$ in $L^1(\mu)$

I've tried to split it up: $$\int_\mathbb{R} |f_ng_n -fg|dx = \int_\mathbb{R} |f_n(g_n-g) -g(f_n-f)|dx$$ The second part goes to zero but what about the first part?

$\endgroup$
1
$\begingroup$

You should try the splitting $$ \int_R |f_ng_n-gf| \le \int_R |f(g_n-g)| +\int_R |(f_n-f)g_n| $$ Edit: Second integral converges since $f_n\to f$ in $L^1$ and $g_n$ is bounded uniformly.

For the first integral: Convergence in measure implies a.e. convergence of a subsequence $g_{n_k}\to g$. This implies $g\le 2$ a.e. The function $f(g_{n_k}-g)$ converges pointwise a.e. to zero, and is dominated by $4|f|\in L^1$. Hence by DCT the first integral vanishes for the particular subsequence.

Then we can prove that for any subsequence there is another subsequence such that the first integral converges to zero on this subsequence. Hence converge $\int |f(g_n-g)|\to 0$ follows.

I am sure, this can be done more elegantly ...

$\endgroup$
6
  • $\begingroup$ but we do not know if f is bounded? $\endgroup$ – Johan Apr 9 '14 at 8:02
  • $\begingroup$ We have $g_{n_k}\to g$ pointwise a.e. for a subsequence. Now we can apply Lebesgue dominated convergence theorem for the first integral. $\endgroup$ – daw Apr 9 '14 at 8:09
  • $\begingroup$ Can you please expand a little how we can use the DCT just because we have a convergent subsequence. $\endgroup$ – Johan Apr 9 '14 at 8:16
  • $\begingroup$ Edited the answer. $\endgroup$ – daw Apr 9 '14 at 8:21
  • $\begingroup$ So we prove that every subsequence has a convergent subsubsequence and thats why $g_n$ converge to $g$? isnt that the same thing as stating that convergence in measure would imply convergence a.e? $\endgroup$ – Johan Apr 9 '14 at 8:37

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.