# The limit of a Lebesgue integral

I'm trying to prove the following exercise：

Suppose that $$f,f_n,g\in\mathcal{L}^2(\mathbb{R})$$, $$n=1,2,...$$, $$f_n$$ converges to $$f$$ $$\mu-$$almost everywhere, and $$\int_{\mathbb{R}}|f_n(x)|^2 dx\leq 1 , n=1,2,...$$ then we have: $$\lim_{n\rightarrow \infty}\int_{\mathbb{R}}|f_n(x)-f(x)||g(x)|dx=0$$ Here are my attempts:

use the Cauchy inequality, we have: $$\left(\int_{\mathbb{R}}|f_n(x)-f(x)||g(x)|dx\right)^2\leq \int_{\mathbb{R}}|f_n(x)-f(x)|^2dx\int_{\mathbb{R}}|g(x)|^2dx$$ and since $$g\in \mathcal{L}^2(\mathbb{R})$$, $$\int_{\mathbb{R}}|g(x)|^2dx<\infty$$, it suffices to prove that: $$\lim_{n\rightarrow \infty}\int_{\mathbb{R}}|f_n(x)-f(x)|^2dx=0$$ but I was stucked there since I can't find a dominating function and apply the Lebesgue Dominated Convergence Theorem. I also tried Vitali Convergence Theorem, but in vain as well. Can anybody give me some hints?

• Your last statement isn't true. $f_n(x) = \sqrt{n} \chi_{[0,1/n]}(x), f(x) = 0, g(x) = \chi_{[0,1]}(x)$ where $\chi_S$ is the characteristic function $1$ if $x \in S$ and $0$ if $x \notin S$. Commented May 6, 2022 at 14:44

What is asked to show is that boundedness in $$\mathbb L^2$$ combined with almost everywhere convergence implies that $$\lvert f_n-f\rvert$$ converges weakly in $$\mathbb L^2$$ to $$0$$. In general, as the example given by aschepler $$f_n(x) = \sqrt{n} \chi_{[0,1/n]}(x), f(x) = 0$$, strong convergence in $$\mathbb L^2$$ does not hold.
1. Replacing $$f_n$$ by $$\lvert f_n-f\rvert$$, it suffices to treat the case $$f=0$$ and $$f_n$$ non-negative.
2. Using the fact that linear combinations of indicator functions of sets of measure $$0$$ are dense in $$\mathbb L^2$$, boundedness of $$\left(\lVert f_n\rVert_{\mathbb L^2}\right)_{n\geqslant 1}$$ and Cauchy-Schwarz inequality, it suffices to show that $$\int f_n(x)\mathbf{1}_A(x)dx\to 0$$ for each set $$A$$ of finite measure.
3. To do so, for a fixed $$\varepsilon$$, let $$B_n=\{x\in\mathbb R\mid f_n(x)>\varepsilon\}$$. Then $$\int f_n(x)\mathbf{1}_A(x)dx=\int f_n(x)\mathbf{1}_A(x)\mathbf{1}_{B_n}(x)dx+\int f_n(x)\mathbf{1}_A(x)\mathbf{1}_{\mathbb R\setminus B_n}(x)dx\leqslant \lVert f_n\rVert_{\mathbb L^2}\lambda\left(A\cap B_n\right)^{1/2}+\varepsilon\lambda(A).$$ To conclude, notice that the first terms is a bounded sequence multiplied by something which converges to $$0$$ hence for each positive $$\varepsilon$$, $$\limsup_{n\to\infty}\int f_n(x)\mathbf{1}_A(x)dx\leqslant \varepsilon\lambda(A).$$