Let $f(x)= \frac{1}{\sqrt{x}}$ for $0 < x < 1$ . I am asked to show that for some enumeration on the rationals,

$$F(x)= \sum_{n=1}^{\infty} 2^{-n} f(x - r_n)$$ is integrable.

$\textbf{My Attempt}$ Let $$F_n (x) = \sum_{k=1}^{n} 2^{-k}f(x-r_k)$$ Then $F_n(x)$ is a monotone increasing sequence of measurable functions which converges to $F(x)$. Thus by the Monotone Convergence Theorem we have that $$ \lim_{n \to \infty} \int_{0}^{1} \sum_{k=1}^{n} 2^{-k}f(x-r_k) = \lim_{n \to \infty} \sum_{k=1}^n 2^{-k} \int_{0}^{1} f(x - r_k)$$

I know that $\int_{0}^{1} f(x)= 2$, when evaluating the improper Riemann integral. How do I translate this to Lebesgue integration to finish my proof?


$f$ is Riemann integrable on $[n^{-1},1]$ for all $n$. So the Lebesgue integral $\int_{n^{-1}}^1 f(x)$ exists, and equals the Riemann integral. Now use the monotone convergence theorem on the sequence $f\cdot\chi_{[n^{-1},1]}$ to get the Lebesgue integral of $f$ over $[0,1]$.


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