Upper bound on the integral of a step function inequality

Question

Problem

I have been working through content in real analysis and came across across an inequality in a proof that is unclear to me.

We are told that a function $\psi$ is a step function on the interval $[a,b]$ where there exists partition $P=${$p_0,…p_k$} for which $\psi$ is constant on all intervals $(p_{i-1},p_i)$.

The proof begins with the inequality:

Note: we define $a_i=a+i\frac{(b-a)}n$

I understand that we can replace each value $\psi (a_i)$ with $\lVert \psi \lVert_\infty$ to bound the left hand side.

However, I’m not quite sure how they have derived the upper bound $2k\frac{(b-a)}n\lVert \psi\lVert_\infty$ from the information provided.

I would be grateful if anyone could shed some light on what I’m failing to understand here.

Edit

I have included more of the proof for clarity where the aim is to show that as n$\rightarrow$$\infty$ the left hand side converges to 0:

where $S[a,b]$ is the set of step functions on $[a,b]$.

Answer 1

The idea is the following. Consider another step function $g(t)$, whose discontinuities are at the points $a_i = a + i\frac{b-a}{n}$ and which has the step heights $g(a_i) = \psi(a_i)$ with a uniform step width $\frac{b-a}{n}$. This function has the integral $$ \int_a^b g(t) \mathrm{d}t = \sum_{i=1}^n \frac{b-a}{n} \psi(a_i). $$ The key observation of the authors is then, that $g(t)$ and $\psi(t)$ may only differ in intervals of size $\frac{b-a}{n}$ around the points where $\psi(t)$ is discontinuous. As $\psi(t)$ has at most $k$ discontinuities and the maximum difference between $g(t)$ and $\psi(t)$ is $2 ||\psi||_\infty$, the difference between the two integrals satisfies. $$ \left| \int_a^b \psi(t)\mathrm{d}t - \int_a^b g(t)\mathrm{d}t \right| \leq k\frac{b-a}{n} 2 ||\psi||_\infty. $$