# Question regarding convergence in the pth mean

Here's what I am trying to prove:

Let $$\Omega = [0,1]$$, $$1. Let $$\{ 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.

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.

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.
$$\left|\int f_ng-\int fg\right|\leq \int|f_n-f||g|\underset{\text{Holder ineq.}}{\leq} \|f_n-f\|_p\|g\|_q\underset{n\to \infty }{\longrightarrow }0.$$
• How does it follow from almost everywhere convergence that $\lVert f_n -f \rVert _p \to 0$ as $n\to \infty$? Oct 30, 2022 at 15:48