Integral of a L^1 function

Question

I have come across this inequality that is supposedly not true for $L^1(0,T)$ functions, but works for $g\in L^p(0,T), \, p>1$: $$\forall \epsilon>0 \;\exists \delta>0 \; \text{s.t.} \;|\int_{t-\delta}^tg(s) ds|<\epsilon$$

How is that so?

Answer 1

Actually, it is true that if $g$ is integrable ($\in L^1([0,T])$) then for any $\epsilon>0$ there exists $\delta > 0$ such that if $S \subset [0,T]$ with Lebesgue measure less than or equal to $\delta$ we have

$$\left|\int_S g(x) \, dx\right| < \epsilon.$$

This can be proved using the fact that $g$ can be approximated by a bounded simple function $\phi$ such that

$$\int_S |g(x) - \phi(x)| \, dx \leqslant \int_{[0,T]} |g(x) - \phi(x)| \, dx < \epsilon/2.$$

Since $\phi$ is bounded we have $|\phi(x)| \leqslant M$ and if $m(S) \leqslant \delta = \epsilon/(2M)$, then

$$\int_S |\phi(x)| \, dx < M m(S) \leqslant \epsilon/2.$$

Thus,

$$\left|\int_S g(x) \, dx\right| \leqslant \int_S |g(x)| \, dx \leqslant \int_S |g(x) - \phi(x)| \, dx + \int_S |\phi(x)| \, dx < \epsilon.$$

In this case $S = [t - \delta,t]$ and $m(S) = \delta.$