Exercise 3.6.1 (Uniform series can be interchanged with integrals)

Question

Use Theorem 3.6.1 to prove Corollary 3.6.2.\par 

	extbf{Theorem 3.6.1.} Let $[a, b]$ be an interval, and for each integer $n \geq 1$, let $f^{(n)}:[a, b] ightarrow \mathbb{R}$ be a Riemann-integrable function. Suppose $f^{(n)}$ converges uniformly on $[a, b]$ to a function $f:[a, b] ightarrow \mathbb{R}$. Then $f$ is also Riemann integrable, and

$$
\lim _{n ightarrow \infty} \int_{[a, b]} f^{(n)}=\int_{[a, b]} f .
$$

	extbf{Corollary 3.6.2.} Let $[a, b]$ be an interval, and let $\left(f^{(n)}ight)_{n=1}^{\infty}$ be a sequence of Riemann integrable functions on $[a, b]$ such that the series $\sum_{n=1}^{\infty} f^{(n)}$ is uniformly convergent. Then we have

$$
\sum_{n=1}^{\infty} \int_{[a, b]} f^{(n)}=\int_{[a, b]} \sum_{n=1}^{\infty} f^{(n)}
$$

Prakash Anand · Accepted Answer

\begin{proof}
Let $g^{(N)}=\sum_{n=1}^{N} f^{(n)}$. So we know that $g^{(N)}$ is Riemann integrable since each $f^{(n)}$ is Riemann integrable.

Also $g^{(N)} ightarrow \sum_{n=1}^{\infty} f^{(n)}=g$, when $N ightarrow \infty$, and convergence is uniformy by assumption. Then from Theorem 14.6.1 we have that

$$
\begin{aligned}
\int_{[a, b]} g &=\lim _{N ightarrow \infty} \int_{[a, b]} g^{(N)} \
&=\lim _{N ightarrow \infty} \int_{[a, b]} \sum_{n=1}^{N} f^{(n)} \
&=\lim _{N ightarrow \infty}\left[\int_{[a, b]} f^{(1)}+\int_{[a, b]} f^{(2)}+\cdots+\int_{[a, b]} f^{(N)}ight] \
&=\int_{[a, b]} f^{(1)}+\int_{[a, b]} f^{(2)}+\cdots \
&=\sum_{n=1}^{\infty} \int_{[a, b]} f^{(n)}
\end{aligned}
$$

So we have $\int_{[a, b]} g=\sum_{n=1}^{\infty} \int_{[a, b]} f^{(n)}$.

But, $\int_{[a, b]} g=\int_{[a, b]} \sum_{n=1}^{\infty} f^{(n)}$

Thus,

$$
\sum_{n=1}^{\infty} \int_{[a, b]} f^{(n)}=\int_{[a, b]} \sum_{n=1}^{\infty} f^{(n)}.
$$
\end{proof}

Exercise 3.6.1 (Uniform series can be interchanged with integrals)

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Exercise 3.6.1 (Uniform series can be interchanged with integrals)

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