If $f_1, f_2, \dots$ are Riemann integrable and $\sum_n \int |f_n| < \infty$ is $f=\sum_n f_n$ Riemann Integrable?

I believe the answer is no, because isn't this sort of scenario the motivation for introducing the Lebesgue integral. I am having trouble synthesizing a counter-example.

  • $\begingroup$ If you consider $f_n(x)=x$ for all $n$, then $\sum \int_{-a}^a f_n = 0$ but $f$ doesn't exist. $\endgroup$ – Dog_69 Feb 28 '18 at 16:33
  • $\begingroup$ @Josué Tonelli-Cueto Ok, thanks, I see where I made an error. I delete my answer $\endgroup$ – Kelenner Feb 28 '18 at 16:44
  • 1
    $\begingroup$ @Kelenner Your answer is easily corrected... $\endgroup$ – man and laptop Feb 28 '18 at 16:45

Let $[a,b]\subseteq\mathbb{R}$ be a an interval and let $\{r_1,r_2,r_3,...\}$ be an enumeration of the rational numbers in the set $\mathbb{Q}\cap[a,b]$. Define $$f_n(x):=\chi_n(x)= \begin{cases} 1 & x=r_n \\ 0 & \text{else} \end{cases} $$ Clearly for each $n$ we have $f_n$ is Riemann integrable with Riemann integral $\int_a^bf_n(x)\,dx=0$. Morevoer $$\sum_n\int_a^b|f_n(x)|dx=\sum_n\int_a^bf_n(x)dx=\sum_n 0=0<+\infty$$ But $f(x):=\sum_nf_n(x)$ (pointwise) is the following function $$f(x):=\sum_n\chi_n(x)= \begin{cases} 1 & x\in\mathbb{Q}\cap[a,b]\\ 0 & \text{else} \end{cases}$$ which is not Riemann integrable.


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