Improper integral and lower Riemann sums

Question

Given $f$ is positive and continuous on $(0,1]$ and its improper integral exists there. Is it true that the lower Riemann sums converges to the integral?

I'm thinking about using definition but reach nowhere.

Answer 1

Yes, the lower sums converge to $\int_0^1 f.$ Let $\epsilon>0.$ Choose $b\in (0,1)$ such that

$$\int_0^b f > \int_0^1 f - \epsilon.$$

Let $P_n= \{x_0, \dots ,x_n\}$ be the uniform partition of $[0,1]$ with $n$ subintervals. Then as $n\to \infty,$

$$L(P_n,f) \ge \sum_{x_k < b}m_k\cdot (1/n) \to \int_0^b f.$$

Now the sum in the middle is not quite a lower sum for $\int_0^b f.$ But we're just one summand short of a partition of $[0,b],$ so it won't matter in the limit. It follows that for large $n,$

$$\int_0^1 f \ge L(P_n,f) > \int_0^1 f - 2\epsilon.$$

This gives the desired result.

Answer 2

Let $\epsilon>0$ Since $\int_0^1f$ is convergent, there is a $a>0$ such that $$ \int_0^a f\le\frac{\epsilon}{2}. $$ Since $f$ is integrable on $[b,1]$, there is a lower sum $S$ such that $$ 0\le\int_a^1f-S\le\frac{\epsilon}{2}. $$ Let $m_a=\inf_{0< x\le b}f(x)$. Then $m_a\ge0$ and $m_a\,a+S$ is a lower sum for $f$ on $(0,1]$. We hace $$ 0\le\int_0^1 f-(m_a\,a+S)=\int_0^af-m_a\,a+\int_a^1f-S\le\int_0^af+\int_a^1f-S\le\epsilon. $$ The same argument can be carried out if $f$ is bounded below. Note that the continuity condition is not used.