A limit of integrals

Question

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 $x$.

score 2 · Accepted Answer · answered May 15 '13 at 08:12

2

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.

answered May 15 '13 at 08:12

Mariano Suárez-Álvarez

135,076

You will have convergence, though, if the function $f$ is increasing, for example. This follows from the monotone convergence theorem. – Mariano Suárez-Álvarez May 15 '13 at 08:18
Thank you very much. I had better take $f$ piecewise continuous. – Driss Alami May 15 '13 at 08:24
Actually, left-continuity should suffice I think, since $h\lfloor \frac{t}{h}\rfloor \leq h$. – fgp May 15 '13 at 08:29
That gives you pointwise convergence, you do need something more usually. – Mariano Suárez-Álvarez May 15 '13 at 08:36

A limit of integrals

1 Answers1