Let $f:[0,T]\to\mathbb{R}$ be a Lebesgue integrable function. For each $h>0$ we define the piecewise function $f_h$ by $$f_h(t)=f(h\left[\frac{t}{h}\right])\quad\mbox{for}\quad t\in[0,T].$$
Can we affirm that $$\lim_{h\to 0}\int_0^Tf_h(t)dt=\int_0^Tf(t)dt ?$$
*N.B :*$\left[ x\right]$ is the entire part of the real $x$.
Let $T=1$ and let $f:[0,1]\to\mathbb R$ be the function which takes the value $0$ on rational numbers and $1$ on the irrational ones.
If $n\in\mathbb N$, then $f_{1/n}$ is the zero function. It follows easily from that that $\liminf\limits_{h\to0}\int_0^1f_h=0$ and therefore your equality cannot be true.