# Show step functions are Riemann integrable

We have $$f: [a,b] \to \mathbb{R}$$ is a step function if there exists a partition $$P=\{x_0, \ldots, x_n \}$$ of $$[a,b]$$ such that $$f$$ is constant on the interval $$[x_i, x_{i+1})$$.

I want to show that this piecewise constant function is Riemann-integrable. I know by Riemann's Criterion, $$f$$ is R-integrable $$\iff$$ $$\forall\ \epsilon > 0$$, there is a partition $$P$$ of $$[a, b]$$ such that $$|U(f, P) - L(f, P)| < \epsilon$$.

Now it's clear that the thing to tackle here are the discontinuities at the end of each 'piece' of the function. Stuck on how to do that exactly, idea probably is to bound each discontinuity with the partition I choose.

Or maybe an alternative proof could use the additivity of integrals, and define the function on each 'piece'.

• Did you make a drawing? – user370967 Dec 15 '18 at 22:45
• Yes, I've seen an example with a single discontinuity. So extending that idea, I suppose you could define a partition $\{x_0, x_0 + M, x_1, x_1 + M, ...\}$, where M captures those 'jumps'. And then with that bound the difference in lower sums and upper sums appropriately if that makes sense. – SS' Dec 15 '18 at 22:53
• Prove that $f$ is integrable on each of $[x_i, x_{i+1}]$ or equivalently the problem is reduced to the case when $f$ is constant on $[a, b)$. – Paramanand Singh Dec 16 '18 at 5:54