Is this intuitive true that $\int_Ef=\int_{I_n}f$

Question

When proving general integral, we usually consider simple function first. For example:

Simple function -->Bounded function-->Non-negative function-->General function

For Lebesgue integral, in the definition of integral of non-negative function we have a concept finite support, which means $m(\{x\in\mathbb{R}|f(x)\neq0\})<\infty$. Assume $E_0=\{x\in\mathbb{R}|f(x)\neq0\}$, then we have $$\int_\mathbb{R}f=\int_{E_0}f$$

In this case, can we always find a bounded interval, say $I_n=[-n,n]$ for which $$\int_{E_0}f=\int_{I_n}f$$ where $n<\infty$.

Try: since $\lim_{n\rightarrow\infty}m(E-[-n,n])=0$, I think my intuitive may be correct?

Answer 1

A function may have a support of finite measure, but the equality $$\tag{}\int_{E_0}f=\int_{I_n}f$$ may fail. For example, define $$f(x)=\sum_{j=1}^{+\infty}\mathbf 1_{\left[j,j+2^{-j}\right]}.$$ In this case, $E_0=\bigcup_{j=1}^{+\infty}\left[j,j+2^{-j}\right]$, whose measure is finite but (*) fails for each $n$, since $$\int_{E_0}f(x)-\int_{I_n}f(x)=\sum_{j=n+1}^{+\infty}2^{-j}\neq 0.$$