TITLE: Question about proof of second mean value theorem for integrals.
QUESTION [1 upvotes]: 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?

REPLY [0 votes]: 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.
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