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.
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$\begingroup$ 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 $f\chi_A$. $\endgroup$ – J.G Jun 20 at 19:32
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$\begingroup$ @user293121 your first statement is false. the inequality can certainly be true even if $f$ is sometimes negative $\endgroup$ – mathworker21 Jun 20 at 19:33
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$\begingroup$ What is mean by point wise dominated ? $\endgroup$ – Ling Min Hao Jun 20 at 19:33
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$\begingroup$ @LingMinHao a function $f$ pointwise dominates a function $g$ if $f(x) \ge g(x)$ for each $x$. $\endgroup$ – mathworker21 Jun 20 at 19:33
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$\begingroup$ @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. $\endgroup$ – J.G Jun 20 at 20:09
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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
$$\int_Bf := \int_Af \chi_B = \sup_{\phi \leqslant f\chi_B}\int_A \phi \leqslant \int_A f$$
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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 $\int_B f\le \int_A f$.
answered Jun 20 at 21:40
