This appears as the 7th exercise of Chapter 10 in Apostol's book of Mathematical Analysis. $\{f_n\}$ is a sequence of functions and $p_n$ is an increasing sequence such that $p_n \rightarrow +\infty$. We have that

  1. The sequence of functions $\{f_n\}$ converges uniformly to $f$ on $[a, b]$ for every $b \ge a$.
  2. Each $f_n$ is Riemann-integrable on $[a, b]$ for every $b \ge a$.
  3. $|f_n(x)| < g(x)$ almost everywhere on $[a, +\infty)$ for some nonegative g, which is improper Riemann integratable on $[a, +\infty)$.

Prove that $f$ and $|f|$ are improper Riemann-integrable on $[a, +\infty)$, the sequence $\{\int_a^{p_n}f_n(x)dx\}$ converges and $$ \int_a^{+\infty}f(x)dx = \lim_{n\rightarrow+\infty}\int_a^{p_n}f_n(x)dx. $$ It is easy to verify that the improper Riemann integral exists, but the convergence of $\{\int_a^{p_n}f_n(x)dx\}$ is frustrating.


Given $\epsilon>0$, since $f,g$ are improper Riemann integrable, choose some $b_{0}>a$ large enough such that $\left|\displaystyle\int_{b}^{\infty}f(x)dx\right|<\epsilon$ and $\displaystyle\int_{b}^{\infty}g(x)dx<\epsilon$ for all $b\geq b_{0}$. Now uniform convergence of $(f_{n})$ on $[a,b_{0}]$ gives some $N$, \begin{align*} \left|\int_{a}^{b_{0}}f_{n}(x)dx-\int_{a}^{b_{0}}f(x)dx\right|<\epsilon,~~~~n\geq N. \end{align*} We may assume also that $p_{n}>b_{0}$ for all such $n$, then \begin{align*} \left|\int_{a}^{p_{n}}f_{n}(x)dx-\int_{a}^{\infty}f(x)dx\right|&\leq\left|\int_{a}^{b_{0}}f_{n}(x)dx-\int_{a}^{b_{0}}f(x)dx\right|\\ &~~~~+\left|\int_{b_{0}}^{p_{n}}f_{n}(x)dx-\int_{b_{0}}^{\infty}f(x)dx\right|\\ &<\epsilon+\int_{b_{0}}^{p_{n}}|f_{n}(x)|dx+\left|\int_{b_{0}}^{\infty}f(x)dx\right|\\ &<2\epsilon+\int_{b_{0}}^{p_{n}}g(x)dx\\ &<2\epsilon+\int_{b_{0}}^{\infty}g(x)dx\\ &<3\epsilon. \end{align*}


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