# Integral of measurable subset of nonnegative measurable function.

Suppose $$f$$ is a nonnegative measurable function on bounded measurable set $$A$$. For $$B\subset A$$ measurable subset, then $$\int_{B}fdm\leq\int_{A}fdm$$. How to prove this ? Is there any particular hint ? (If I use the definition of supremum , I still need to prove $$\int_{B}f_1dm\leq\int_{A}f_1dm$$ for simple function $$f_1\leq f$$ for which I don't know how to prove it rigorously.

• This isn't true unless $f\geq 0$...assuming this though, this follows by just noting that any simple function that is pointwise dominated by $f\chi_B$ is also pointwise dominated by $f\chi_A$. – J.G Jun 20 at 19:32
• @user293121 your first statement is false. the inequality can certainly be true even if $f$ is sometimes negative – mathworker21 Jun 20 at 19:33
• What is mean by point wise dominated ? – Ling Min Hao Jun 20 at 19:33
• @LingMinHao a function $f$ pointwise dominates a function $g$ if $f(x) \ge g(x)$ for each $x$. – mathworker21 Jun 20 at 19:33
• @mathworker21 it doesn't matter that the inequality can sometimes be true, the general statement that was to proved requires it be true for all $B\subseteq A$, which fails unless $f\geq 0$ a.e. – J.G Jun 20 at 20:09

Suppose $$\phi$$ is a simple function such that $$0 \leqslant \phi \leqslant f \chi_B$$. Since $$f \chi_B \leqslant f$$ on $$A$$ it follows that $$\phi \leqslant f$$ on $$A$$, and
$$\int_A \phi \leqslant \sup_{\phi \leqslant f }\int_A \phi =\int_Af$$
Thus, $$\int_Af$$ is an upper bound for $$\int_A \phi$$ for every $$\phi \leqslant f \chi_B$$, and
$$\int_Bf := \int_Af \chi_B = \sup_{\phi \leqslant f\chi_B}\int_A \phi \leqslant \int_A f$$
Let $$g(x)=f(x)$$ for $$x$$ in $$B$$ and $$g(x)=0$$ otherwise. Then $$\int_A g=\int_B f$$. However $$g\le f$$ on $$A$$, so $$\int_A g\le \int_A f$$. Therefore $$\int_B f\le \int_A f$$.