Show $\lim\limits_{h\to 0^+}h \sum_{n=1}^{\infty}f(nh)=\int_0^{\infty}f(x)dx$

Question

Let $f$ be a monotonous function over $[0, \infty)$ for which the improper integral $\int_0^{\infty}f(x)dx$ converges. I want to show $$\lim\limits_{h\to 0^+}h \sum_{n=1}^{\infty}f(nh)=\int_0^{\infty}f(x)dx$$ First, we notice that for the integral to converge $f$ has to be decreasing. Now we have $$\int_{(n-1)h}^{nh}f(x)dx\geq f(nh)h\geq\int_{nh}^{(n+1)h}f(x)dx$$ Summing from $n=1$, we get $$\sum_{n=1}^{\infty}\int_{(n-1)h}^{nh}f(x)dx\geq \sum_{n=1}^{\infty}f(nh)h\geq \sum_{n=1}^{\infty}\int_{nh}^{(n+1)h}f(x)dx$$ $$\int_{0}^{\infty}f(x)dx\geq h\sum_{n=1}^{\infty}f(nh)\geq \int_{h}^{\infty}f(x)dx$$ We apply the limit $$\lim\limits_{h\to 0^+}\int_{0}^{\infty}f(x)dx\geq \lim\limits_{h\to 0^+}h\sum_{n=1}^{\infty}f(nh)\geq \lim\limits_{h\to 0^+}\int_{h}^{\infty}f(x)dx$$ $$\int_{0}^{\infty}f(x)dx\geq \lim\limits_{h\to 0^+} h\sum_{n=1}^{\infty}f(nh)\geq \int_{0}^{\infty}f(x)dx$$ Finally $$\lim\limits_{h\to 0^+} h\sum_{n=1}^{\infty}f(nh)= \int_{0}^{\infty}f(x)dx$$ Is this line of reasoning correct?