# Proving that:$\int_X f_n g \, d\mu \to \int_X fg \, d\mu$ for all $g$ in $\mathscr{L}^q (X)$

I found the following exercise and I'd like to know if my answer is correct.

Let $(X, \mathscr A, \mu)$ a finite measure space. Let $\{f_n\}$ a sequence of measurable functions such that $\|f_n\|_p\le M$ for a real constant $M$ $(1<p<\infty)$ and suppose that $f_n \xrightarrow{\text{a.e.}} f$. Then $$\int_X f_n g \,d\mu \to \int_X fg \,d\mu$$ for all $g$ in $\mathscr{L}^q (X)$, where $q^{-1}=1- p^{-1}$.

Since $\int |f_n|^p \,d\mu\le M^p$ by Fatou's lemma it follows that $\int |f|^p \,d\mu\le M^p<\infty$, so $f$ belongs to $\mathscr{L}^p (X)$. Let $g$ in $\mathscr{L}^q$ arbitrary but fixed.

Then by Hölder inequality it follows that $$\|f_n g\|_1 \le \|f_n\|_p\|g\|_q\le M \|g\|_q<\infty$$ and $$\|f g\|_1 \le \|f\|_p\|g\|_q \le M \|g\|_q<\infty$$ thus $f_n g$ and $fg$ belongs to $\mathscr L ^1(X)$. Now since $g$ is in $\mathscr L ^q (X)$, then $g$ is finite a.e.,without loss of generality we may assume that $g$ is real-valued so $f_n g \xrightarrow{ a.e} fg$.

Let $\nu (A) = \int_A |g|^q \,d\mu$, so $\nu$ is absolutely continuous with respect to $\mu$, also $\nu$ is a finite measure on $(X, \mathscr A )$. We can use the $\epsilon-\delta$ definition of absolutely continuous, so given $\epsilon>0$ there is a $\delta>0$ such that for $\mu(A)<\delta$ then $\nu (A) <\epsilon^q$ for any $A$ in $\mathscr A$.

Now by Egoroff's thm exists a $B$ in $\mathscr A$ such that $(f_ng) (x)\xrightarrow{\text{uniformly}} (fg) (x)$ for $x$ in $B$ and $\mu(X\setminus B) <\delta$. Let $N$ such that for all $n\ge N$, $|(f_ng)(x)-(fg)(x)|<\epsilon$ for all $x\in B$. Thus

\begin{align*}\left |\int_X (f_n-f) g \, d\mu \right|&\le \int_B |f_ng-fg| \, d\mu +\int_{X-B}|f_n g|\, d\mu +\int_{X-B}|fg|\,d\mu \\[6pt] &\le \int_B |f_ng-fg| d\mu + 2M \left( \int_{X-B}|g|^qd\mu \right)^{1/q}\\[6pt] &\le\epsilon \mu(X)+2M (\nu(X\setminus B))^{1/q}\\[6pt] &\le \epsilon (\mu(X)+2M)\end{align*}

Since $\mu(x)<\infty$ and $M<\infty$ the result follows.

• Looks good to me! – GaussTheBauss Dec 24 '15 at 19:04
• @GaussTheBauss I think there is a mistake because in $|f_n|\le M$ because we only have $\|f_n\|\le M$ – Jose Antonio Dec 24 '15 at 19:46
• You're right. Here's how you can fix it. Since $f_n g$ and $fg$ are integrable, then so is $k_n:=f_n g-fg$. Thus the integral of $k_n$ over a small set is small (you have used that fact already in your proof - absolute continuity). Here your small set is $X \setminus B$, which can be as small as you wish from Egoroff's Theorem. What do you think? – GaussTheBauss Dec 24 '15 at 19:58
• @GaussTheBauss I made a small change using for $\nu (A)= \int_A |f|^q d\mu$ and also using that $|fg|_1\le M\|g\|_q$ and t $|f_ng|_1\le M\|g\|_q$ and the argument follows almost in the same way. What do you think? (Let me think about your idea). Thank you so much :) – Jose Antonio Dec 24 '15 at 20:03
• I like what you did – GaussTheBauss Dec 24 '15 at 20:23

Your answer is correct but since $\mu(X)<\infty$, here is a short way to solve it using Lebesgue dominated theorem.

Let fix $g\in\mathscr{L}^q (X)$ then $g\in \mathscr{L}^1 (X)$ indeed, by Hölder inequality we have,

$$\int_X 1\cdot |g|d\mu\le \left(\int_X 1d\mu\right)^{1/p} \left(\int_X |g|^qd\mu\right)^{1/q} =\mu\left(X\right)^{1/p}\|g\|_q<\infty$$

1-since $\lim\limits_{n\to\infty} f_n(x)=f(x)$for $\mu-$almost every $x\in X$ $$\lim_{n\to\infty} f_n(x)g(x) =f(x) g(x)~~~a.e$$

2-for all n we have, $$|f_ng|\le M|g|~~a.e$$ and $g\in \mathscr{L}^1 (X).$

It springs from Lebesgue convergence dominated theorem that

$$\lim_{n\to\infty}\int_X f_n g \,d\mu =\int_X fg \,d\mu$$