Exercise 1.4.41 (Uniform convergence and integrals)

Question

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ewcommand{\intx}[1]{\int_{X} #1 \mathop{}\!\mathrm{d}{\mu}}$
Let $(X,\mathcal{B},\mu)$ be a finite measure space, and let $\left(fight)_{n\in\mathbb{N}}$ be a sequence of absolutely integrable functions from $X$ to $\mathcal{C}$ that converge uniformly to a limit $f:X	o\mathbb{C}$. Show that $$\lim_{n	o\infty} \intx{f_n} = \intx{\lim_{n	o\infty} f_n}.$$

Elu Thingol · Accepted Answer

Let $f:=\lim_{n	o\infty}f_n$. By definition we have
$$\forall \epsilon >0 \quad \exists N\in\mathbb{N} \quad \forall n > N: \quad d_{\infty}(f_n,f)\leq \epsilon.$$
Then $f$ is measurable (Exercise 1.4.29), and since the uniform convergence of the functions implies the uniform convergence of the norms by the triangle inequality
$$d_{\infty}\left(|f_n(x)|,|f(x)|ight) = \sup\left\{\left||f_n(x)|-|f(x)|ight|: x \in X ight\} \leq \sup\left\{\left|f_n(x)-f(x)ight|: x \in X ight\} =  d_\infty(f_n,f) \leq \epsilon$$
we see that $|f| = \lim_{n	o\infty}|f_n|$ is absolutely integrable as well
$$\int_{X} |f| \,\mathrm{d}{\mu} = \intx{|f|-|f_n|} + \intx{|f_n|} \leq \intx{\epsilon} + \intx{f_n} = \epsilon\mu(X) + \intx{|f_n|} < \infty.$$
Now we show the convergence. For the $\epsilon/\mu(X)$ we can choose $n\geq N$ so that we have
$$\left|\intx{f_n} - \int_{X} f \,\mathrm{d}{\mu} ight| = \left|\intx{(f_n-f)} ight| \leq \intx{\frac{\epsilon}{\mu(X)}} = \epsilon.$$
as desired.

Exercise 1.4.41 (Uniform convergence and integrals)

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Exercise 1.4.41 (Uniform convergence and integrals)

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