Question about proof of second mean value theorem for integrals.

Question

I am reading about second mean value theorem and have questions about some parts of the proof

Here $f$ is decreasing and positive function on $[a,b]$ and $g$ is integrable on $[a,b]$

$p=\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}}[f(x)-f(x_i)]g(x)dx$

Author says because $g$ is bounded $|g(x)|\leq L$

Therefore

$|p|$ $\leq$ $\sum_{i=0}^{n-1}\int_{x_i}^{x_{i+1}}|f(x)-f(x_i)||g(x)|dx$ $\leq$$L\sum_{i=0}^{n-1}w_i(f)\Delta x_i$

where $w_i(f)$ is oscilation of function on $i$-th interval.

Can you in detail explain how last inequality is derived?

Answer 1

Recall the definition of the oscillation of $f$ on an interval and its equivalent form

$$\omega_i(f) := \sup_{t\in [x_i,x_{i+1}]}f(t) - \inf_{t\in [x_i,x_{i+1}]}f(t) = \sup_{s,t\in [x_i,x_{i+1}]}|f(s) - f(t)|,$$

Thus, $|f(x) - f(x_i)| \leqslant \omega_i(f)$ for all $x \in [x_i,x_{i+1}]$, and

$$\int_{x_i}^{x_{i+1}}|f(x)-f(x_i)| \, dx \leqslant \omega_i(f)(x_{i+1}-x_i)= \omega_i(f) \Delta x_i$$

That along with $|g(x)| \leqslant L$ yields your last inequality.