# $f_n \rightarrow f$ in measure and $g$ be measurable and finite a.e. Prove that $f_ng \rightarrow fg$ in measure

Let $E$ be of finite measure, $f_n \rightarrow f$ in measure and $g$ be measurable and finite a.e. Prove that $f_ng \rightarrow fg$ in measure

My attempt : We have $g$ is finite a.e thus $D:= \lbrace x : \vert g(x) \vert = \infty\rbrace$ then $\mu D =0$ Now as $f_n$ converges in measure we have following for any $\epsilon$

$E':= \lbrace x : \vert f_n(x)-f(x) \vert \geq \epsilon \rbrace$ then $\mu E\rightarrow 0$ and $n \rightarrow 0$

Now, For any $\epsilon > 0$

$E_0 := \lbrace x : \vert f_n(x)g(x)-f(x)g(x) \vert \geq \epsilon \rbrace = \lbrace x : \vert g(x)\vert \vert f_n(x)-f(x) \vert \geq \epsilon \rbrace \cup D$

but measure of both the sets on RHS is $0$ as $n \rightarrow \infty$ thus $\mu E_0 \rightarrow 0$ and $n \rightarrow 0$

I am not sure if the equation of union of sets which i am trying to formulate is right.

$g$ is finite a.e. thus as you are saying, $D:=\{ x:|g(x)|=\infty\} \implies \mu D=0$. Further, $E' := \{ x : |f_n(x)-f(x)| \geq \epsilon \} \implies \mu E' \rightarrow 0$ if $n \rightarrow \infty$.

Now, I say the following : for any $M>0, M \in \mathbb{Z}$, let $F_M = \{ x:|g(x)| \geq M\}$. Note that by the Borel-Cantelli Lemma, $\displaystyle\lim_{M \rightarrow \infty} \mu F_M \rightarrow 0$. SO just pick $M$, such that $\mu F_M$ is very very small.

Finally, we have that for any $\epsilon > 0$,$$\mu\{x : |f_n(x)g(x) - f(x)g(x)| \geq \epsilon \} = \mu \{x :|f_n(x)-f(x)||g(x)| \geq \epsilon\} \\ \subset \{x : |g(x)| \geq M\} \cup \{ x : |f_n(x) -f(x)| \geq \epsilon/M\}$$.

Now pick $n$ large enough, and because we managed to fix the $M$, we are done.

You were wrong in saying that the RHS goes to zero directly. This is the right way to do it, by using the Borel Cantelli Lemma wisely. Please ask if you do not know the Borel Cantelli Lemma, but anyway it states that given a decreasing sequence of measurable sets $F_M$, such that the measure of the first one is finite $\mu F_1 < \mu E < \infty$, then $\lim_{M \rightarrow \infty} \mu(F_M) = \mu (F_\infty)$, which in our case is $D$, but $\mu D=0$.

• RHS should also be $\geq \epsilon$ right ? Am I missing something ? – Rusty Mar 13 '16 at 6:40
• just give me a few minutes. I will refine my argument because I see flaws. – Teresa Lisbon Mar 13 '16 at 6:46
• Oh yes. This is just perfect. I know Borel- Cantelli and had realised need to do something like you did but was unable to formalize. Thanks. – Rusty Mar 13 '16 at 6:48
• Sorry,but I needed to plug a few holes. Here's an exercise for you if you want: even if $g_n \rightarrow g$, then $f_ng_n \rightarrow fg$! It's just a repetition with a little more care. Have a try. – Teresa Lisbon Mar 13 '16 at 6:57