Let $f\in L^1(\mathbb{R}^n)$. I want to prove that when $m(B-B’)\to 0$ $$\left|\int_{B’}f(x)dx-\int_{B}f(x)dx\right|\to 0.$$
In Riemann integral, if $f$ is absolutely integrable on $[a,b]$, then $|f|$ in bounded, namely $|f(x)|\le M$, on $[a,b]$, and hence we have $$\left|\int_{B’}f(x)dx-\int_{B}f(x)dx\right|= \left|\int_{B’-B}f(x)dx\right|\le \int_{B’-B}|f(x)|dx\le M\cdot m(B’-B),$$ where $B\subset B’$. If $m(B-B’)\to 0$, then we will have $$\left|\int_{B’}f(x)dx-\int_{B}f(x)dx\right|\to 0.$$
But for Lebesgue integral, we know $|f|$ is not necessarily bounded and it’s only finite almost everywhere. In this case we cannot find the supremum; in fact, we even cannot necessarily find the essential supremum. For example, $$f(x):=\left\{\begin{array}{l} \dfrac{1}{\sqrt{|x|}},~~x\in[-1,0)\cup(0,1],\\ +\infty,~~x=0,\\ \dfrac{1}{x^2},~~x\in(-\infty,-1)\cup(1,+\infty). \end{array}\right.$$
We can easily check that $f\in L^1(\mathbb{R})$ and of course $|f|$ is integrable on $[-1,1]$. But now we can see that $|f|$ is not bounded and even the essential supremum is $+\infty$. Now we cannot prove that in the same way.
How to do that? The claim seems to be correct but I don’t know how to prove it. Any help is appreciated. Thanks!
