# for each function $g \in L^2(X, \mu)$, $\int fgd\mu=\lim_{n\to \infty}\int f_ngd\mu$

Let $$(X,\mathcal{A}, \mu)$$ be a finite measure space and let $$\{f_n\}$$ be a sequence of real-valued measurable functions on $$X$$. Suppose that there is a constant $$M$$ such that $$|f_n(x)| \leq M$$ for all $$n$$ and $$x \in X$$. Suppose also that the sequence $$f_n(x)$$ converges almost everywhere to a function $$f$$. Show that $$f$$ is measurable and for each function $$g \in L^2(X, \mu)$$. $$\int fgd\mu=\lim_{n\to \infty}\int f_ngd\mu$$

can I have sth like a sketch of what should I do on rest of this problem?

Is the following enough to show that f is measurable? $$\forall \alpha \in \mathbb{R} , \lbrace x\in X:f(x)>\alpha \rbrace = \bigcup_{k\ge 1}\bigcup_{N\ge 1}\bigcap_{n\ge N}\bigg\lbrace x\in X: f_n(x)\ge \alpha + \frac{1}{k} \bigg\rbrace$$

Since $$f_n \to f$$ we also have that $$|f(x)| \leq M$$

so $$f_n,f$$ are integrable since we are in a finite measure space.

And $$\int|f_n-f| \to 0$$ by hypothesis and Dominated convergence.

Thus $$\left |\int f_ng-\int fg\right | \leq \int|g||f_n-f| \leq ||f_n-f||_2||g||_2 \to 0$$ by Cauchy-Schwartz

and from the fact that $$||f_n-f||_2 \leq 2M||f_n-f||_1 \to 0$$

For the measurability note that $$f(x)=\limsup_nf_n(x)$$ and we know that $$\limsup_nf_n$$ is always measurable if $$f_n$$ are.

• the fact $\|f_n-f\|\to 0$, is a corollary that comes from that $|f_n|\leq M$ and $f_n\to f$ a.e., right? – domath Oct 30 '19 at 19:13
• @stat_yale yes because also $f_n,f$ are integrable since we are in a finite measure space,so it is true by the dominated convergence theorem. – Marios Gretsas Oct 30 '19 at 19:14
• Is it viable to show the $L^2$-concergence of $(f_n)$ and use the continuity of the $L^2$ inner product instead? – Botond Oct 30 '19 at 19:24
• @Botond i showed the L^2 convergence..and use the continuity of the inner product..if i understood your question correctly – Marios Gretsas Oct 30 '19 at 19:27
• My idea was to prove that $(f_n)$ converges in $L^2$ to $f$, and then use the continuity of the inner product to show that $(f_n|\bar{g}) \to (f|\bar{g})$. But I realized that it's the same argument as yours, because the continuity of the inner product need the CS. – Botond Oct 30 '19 at 19:31