Recall that $L^{1}(\mathbb{R})$ is defined as the completion of the Riemann integrable functions on the real line with respect to the norm $||f|| = \int |f|$.
Let $h_n(x)$ be the real-valued functions which is equal to $0$ on $[0, 1-\frac{1}{2^n})$, $1$ on $[1-\frac{1}{2^n}, 1)$ and periodic with period $1$. Let $\chi_{[0,1]}$ be the characteristic function of the interval $[0,1]$.
Show that the infinite series $$ f(x) := \sum_{n=1}^\infty h_n (2^{n(n-1)/2} x) \chi_{[0,1]}(x) $$ can be regarded as a limit of a Cauchy sequence in $L^{1}(\mathbb{R})$.
My attempt: It seems to be helpful to calculate $||h_n (2^{n(n-1)/2} x) \chi_{[0,1]}(x)||$ first.
$$ ||h_n (2^{n(n-1)/2} x) \chi_{[0,1]}(x)|| \\ = \int_{0}^{1} |h_n (2^{n(n-1)/2} x)| dx $$
Let $u = 2^{n(n-1)/2} x$, we have
$$ \int_{0}^{1} |h_n (2^{n(n-1)/2} x)| dx = 2^{-n(n-1)/2}\int_{0}^{2^{n(n-1)/2}}|h_n(u)|du $$
From the property of $h_n(x)$, I should have $\int_{0}^{1}h_n(x)dx = 1/2^n$. But I am stuck at how to proceed given this fact and also how do I know that $f(x)$ is the limit of some Cauchy sequence in $L^1(\mathbb{R})$ after I evaluated the norm.