# Approximate the integral of left-continuous functions?

Assume that you have a continuous function $f:[0,t]\rightarrow \mathbb{R}$. Then it is bounded and Borel-measurable, hence integrable. And using uniform-continuity it can be proved that if you have a sequence of partitions of $[0,t]$, where for each n, $\{t_i^n\}$ is a partition of $[0,t]$, and the norm of the partition goes to 0 as n goes to infinity. Then $\Sigma_{i=0}^{N(n)-1} f(t_i^n)(t_{i+1}^n-t_i^n)$ converges to $\int_o^tf(s)ds$ as n goes to infinity.

But what if we only assume that f is left-continuous and integrable(or bounded if that has an impact on the answer?). Does it then follow that we can approximate the integral by sums of the form above? Can we always do this for any sequence, and do we have to take care of how we choose the evaluation points in each partition?

When trying to prove this I get stuck, because I do not have the uniform continuity as I did above. Intuitively I think it should work if I instead use the sum $\Sigma_{i=1}^{N(n)} f(t_i^n)(t_{i}^n-t_{i-1}^n)$, because of the left continuity. But since I do not have uniform continuity I don't see how to show that this converges.

Any hints?

Update An easier way to ask the question is this: If f is bounded and left continuous, then by DCT $\Sigma_{i=0}^{N(n)-1}f(t_i^n)(t_{i+1}^n-t_i^n)$ converges to $\int_0^tf(s)ds$ as n goes to infinity. Do we have the same type of convergence if f is not bounded, but integrable?

According to If $f:\mathbb{R}\to\mathbb{R}$ is a left continuous function can the set of discontinuous points of $f$ have positive Lebesgue measure?, a left-continuous function is continuous except at countably many points. In particular, this function is almost everywhere continuous, so it is Riemann integrable.
If you drop boundedness, this need not hold: an unbounded integrable left-continuous function can fail to be Cauchy integrable. Consider the interval $[-1,1]$ and the function $f(x) = \begin{cases} x^{-1/2}, & x > 0 \\ 0 , & x \le 0 \end{cases}$. For each $n$, choose a partition $\{t_i^n\}$ that contains $\frac{1}{n^4}$ and $\frac{1}{n^4} + \frac{1}{n}$ as successive points, filling it out with other points less than $\frac{1}{n}$ apart. Then the left-endpoint Riemann sum using this partition is at least $n$, so the sums diverge. Using right endpoints doesn't help either; in that case use $\frac{1}{n^4} - \frac{1}{n}$ and $\frac{1}{n^4}$ as successive points.