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

Question

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.

Answer 1

Your answer is correct but since $\mu(X)<\infty$ But that is a typical consequence of Vitali's convergence Theorem. Here is the generalization.

Theorem Assume $\mu(X)<\infty$ and $1<p<\infty$.

Let $(u_n)_n$ be a bounded sequence of $L^p(X)$ converging almost everywhere to $u$. Then $(u_n)_n$ converges to $u$ in $L^r(X)$ for all $1\leq r<p$.

Proof

The $r$-tightness $(u_n)_n$ obviously holds true since $\mu(X)<\infty$.

The $r$-uniform integrability follows from Hölder inequality.

\begin{align*} \sup_{n\geq 1} \int_E |u_n(x)|^r\,d \mu(x)&\leq \mu(E)^{1-\frac{r}{p}} \sup_{n\geq 1} \big(\int_X |u_n(x)|^p\,d \mu(x)\big)^{r/p}\\ &\leq C \mu(E)^{1-\frac{r}{p}}\to0. \end{align*} as $\mu(E)\to 0$.