Prove that $\int_X f_n \, d\mu \to \int_X f \, d\mu$

Question

Let $(X, \mathcal{A}, \mu)$ be a finite measure space. Let $\{f_n\}_{n=1}^\infty \subset L^2(\mu)$ be a sequence of functions such that $f_n \to f$ almost everywhere (a.e.) and $\|f_n\|_2 \leq M$ for all $n \in \mathbb{N}$. Prove that $\int_X f_n \, d\mu \to \int_X f \, d\mu.$

I always feel like I miss some small but important detail so I would be really grateful if anyone could verify and point any mistakes.

Firstly I note that $L^2$ is closed so the limit $f$ is also in $L^2$. The fact that $\mu(X) < \infty$ and that $f_n,f$ are in $L^2$ imply that $f_n,f$ are integrable. We can use Egorov's theorem to find $E$ such that $f_n\to f$ uniformly on E.$\int_X |f_n-f|d\mu = \int_E |f_n-f|d\mu + \int_{X\setminus E} 1\cdot |f_n-f|d\mu \le \epsilon \mu(E)+(\mu(X\setminus E))^{\frac{1}{2}}(\int_{X\setminus E}(f_n-f)^{2})^{\frac{1}{2}} \le\epsilon\mu(X)+\sqrt{\epsilon}2M$.

As $\mu(X)$ is finite this end the proof. Penultimate inequality is Cauchy-Schwarz.

Answer 1

Taking into account all that was said in comments:

• 1. $L^2$ isn't closed under a.e convergence, as $f_n(x)=\min(n,\frac{1}{\sqrt x})$ on $(0,1]$ with usual measure isn't going to converge to an $L^2$ function. Instead, to obtain the fact that $f\in L^2$, one can go through Fatou's lemma:

$$\int_X f^2d\mu = \int_X \liminf_{n\to \infty}((f_n)^2)d\mu \leq \liminf_{n\to \infty}\int_X (f_n)^2d\mu\leq M^2 < \infty.$$

Thus $f\in L^2$ and $\|f\|_2\leq M$.

• 2. Don't forget to mention that for any $\varepsilon > 0$, there exists $E$ (that depends on $\varepsilon$) such that $\mu(X\setminus E) \leq \varepsilon$ and $f_n\xrightarrow{uniformly} f$ on $E$. Let $N\in \mathbb N$ such that $\forall n\geq N$, $\sup_{x\in E}|f_n(x)-f(x)| \leq \varepsilon.$ Then applying Cauchy-Schwarz with the powers correctly written to $n\geq N$:

$$\begin{align}\int_X |f_n-f|d\mu &= \int_E |f_n-f|d\mu + \int_{X\setminus E} |f_n-f|d\mu \\&\le \varepsilon \mu(E)+\int_{X\setminus E} 1\cdot |f_n|d\mu+\int_{X\setminus E} 1\cdot |f|d\mu\\&\le \varepsilon \mu(E)+(\mu(X\setminus E))^{\frac{1}{2}}\left[\left(\int_{X\setminus E}(f_n)^{2}d\mu\right)^{\frac{1}{2}}+\left(\int_{X\setminus E}f^{2}d\mu\right)^{\frac{1}{2}}\right] \\&\le\varepsilon\mu(X)+2M\sqrt{\varepsilon}.\end{align}$$ The rest will be trivial.

Edit: your way is also fine if you write that $$\left(\int_{X\setminus E}(f_n-f)^{2}d\mu\right)^{\frac{1}{2}} \leq \|f_n-f\|_2 \leq \|f\|_2 + \|f_n\|_2\leq 2M$$ by triangular inequality of the $2$-norm.

Answer 2

This can be obtain by Vitali's convergence theorem (see here for a recent posting) . Notice that for any $n\in\mathbb{N}$ $$\|f_n\|_1\leq\|f_n\|_2\sqrt{\mu(X))}\leq M\sqrt{\mu(X))}$$ Hence $\sup_n\|f_n\|_1<\infty$. Also, for any $\varepsilon>0$ let $\delta=\varepsilon^2/M^2$ so that whenever $\mu(A)<\delta$, $$\int_A|f_n|\,\mu\leq\|f_n\|_2\sqrt{\mu(A)}< \varepsilon$$ To recap, we have that

1. $f_n\rightarrow f$ $\mu$-a.s.
2. $(f_n:n\in\mathbb{N})$ is uniformly integrable.

Then $f_n$ converges to $f$ in $L_1$, i.e. $\|f_n-f\|_1\xrightarrow{n\rightarrow\infty}0$. In particular, $\int_Xf_n\xrightarrow{n\rightarrow\infty}\int_Xf$.