Integral of measurable subset of nonnegative measurable function.

Question

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.

Answer 1

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

$$\int_Bf := \int_Af \chi_B = \sup_{\phi \leqslant f\chi_B}\int_A \phi \leqslant \int_A f$$

Answer 2

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 $\int_B f\le \int_A f$.