# convergence in $L^1$ and convergence $\mu$ a.e imply product convergence in $L^1$

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?

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 ...

• but we do not know if f is bounded? Apr 9, 2014 at 8:02
• 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.
– daw
Apr 9, 2014 at 8:09
• Can you please expand a little how we can use the DCT just because we have a convergent subsequence. Apr 9, 2014 at 8:16
• 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? Apr 9, 2014 at 8:37