Question regarding convergence in the pth mean

Question

Here's what I am trying to prove:

Let $\Omega = [0,1]$, $1<p <\infty$. Let $\{ f_n \}$ be a sequence in $L^p [0,1]$ such that $f_n \to f$ almost everywhere and $f \in L^p [0,1]$. Suppose that there is some $M \in \mathbb R$ such that $\lVert f_n \rVert _p \le M$ for all $n$. Prove that for $g\in L^q [0,1]$ where $1/p + 1/q =1$, we have $\lim \int f_n g d\lambda = \int fg d\lambda$. Here $\lambda$ is the Lebesgue measure.

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Here's my poor attempt:

Let $\{ f_n \}$ be a sequence of functions in $L_p [0,1]$. Since $\lVert \cdot \rVert _p$ is a continuous function on $L^p [0,1]$ and $\lVert f_n \rVert _p \le M$ for all $n$, we have that $\lVert f \rVert _p \le M$.

If we try to estimate $\lvert \int f_n g d\lambda - \int fg d\lambda \rvert \le \lVert f_n -f \rVert _p \lVert g \rVert _q$. If we could somehow how that $\lVert f_n - f \rVert _p \to 0$ as $n \to \infty$, we will be done.

There are certain things that I observe: we have a finite measure space and so almost everywhere convergence implies convergence in measure. However, convergence in measure will not possibly imply convergence in the $p$th mean. So we are hopeless at this certain point. However, I notice that I am not using the fact that $\lVert f_n \rVert \le M$ and $\lVert f \rVert \le M$ for each $n \in \mathbb N$. I do not see how to use it as well.

I am looking for hints that could possibly lead me to a solution to this problem. Any series of hints will be appreciated.

Answer 1

Since $f_{n}\rightarrow f$ a.e., from Egoroff's theorem, for every $\delta>0$ there exists $E\subset\Omega$ such that $m(E^{c})<\delta$ and $f_{n}\rightarrow f$ uniformly on $E$. So for any $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\lvert f_{n}-f\rvert<\dfrac{\epsilon^{p}}{m(E)}$ whenever $n>N$. Hence, $\displaystyle\int_{E}\lvert f_{n}-f\rvert^{p}<\epsilon^{p},\ n>N$. Then since $g\in L^{q}$, from the absolute continuity of integral we have $\displaystyle\int_{E^{c}}\lvert g\rvert^{q}<\epsilon^{q}$. Thus, when $n>N$, we have$$\left| \int f_{n}g-\int fg\right| \leq\int\lvert f_{n}-f\rvert\lvert g\rvert=\left( \int_{E}+\int_{E^{c}}\right) \lvert f_{n}-f\rvert\lvert g\rvert\leq\left( \int_{E}\lvert f_{n}-f\rvert^{p}\right) ^{\frac{1}{p}}\lVert g\rVert_{q}+\lVert f_{n}-f\rVert_{p}\left( \int_{E^{c}}\lvert g\rvert^{q}\right) ^{\frac{1}{q}}\leq\lVert g\rVert_{q}\epsilon+2M\epsilon.$$Since $\epsilon$ is arbitrary, we are done.

Answer 2

You simply have

$$\left|\int f_ng-\int fg\right|\leq \int|f_n-f||g|\underset{\text{Holder ineq.}}{\leq} \|f_n-f\|_p\|g\|_q\underset{n\to \infty }{\longrightarrow }0.$$

The claim follow.