Enumerate the rationals in $[0,1]$ by $\{q_k\}_{k\in \mathbb{N}}$. Define for each $k$, $f_k(x) = 1$ if $q_k I WTS $F(x) = \sum \frac{f_k(x)}{2^k}$ is Riemann integrable on $[0,1]$. I can't find an easy uniform convergence argument. Also, I'm having trouble calculating what the values of $F(x)$ even are. This is obviously my biggest issue. I'd appreciate some help.
Riemann-integrability of function with countably many discontinuities
1
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real-analysis
riemann-integration
1 Answers
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Hint:
$$\left| \frac{f_k(x)}{2^k} \right| \leqslant \frac{1}{2^k}, \\ \sum_{k=1}^\infty \frac{1}{2^k} = 1, $$
and $\sum \frac{f_k(x)}{2^k}$ is uniformly convergent by the Weierstrass test.