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$f$ is Lebesgue integrable over $A$, and $B$ is a measurable subset of $A$. I want to show $$\int_B f=\int_Af\chi_B$$, where $\chi_B$ is the characteristic function of $B$ (it is 1 on B and 0 otherwise). My definition of integrable is "$f$ is integrable if the functions $f^+, f^-$ are integrable."

What I have tried to do is to write $\int_A f\chi_B=\int_Af\chi_B+\int_{A-B}f\chi_B,$ and it would simply follow. However, it is not obvious that I can write that since it is not given that $f$ is bounded. A hint would be helpful.

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1 Answer 1

WLOG, consider the case when $f$ is nonnegative, as we can extend to the general case by writing $f$ as the difference of two nonnegative functions.

$$ \int_{B} f = \int_{B} f^+ - \int_{B} f^- = \int_{A} f^+ \chi_B - \int_{A} f^- \chi_B = \int_A f\chi_B$$

The definition of the Lebesgue integral of $f$ is the supremum of the integrals of simple functions below $f$. Construct a correspondence between the simple functions defined on $B$ below $f$ and the simple functions defined on $A$ below $f \chi_B$. If you can show that a simple function of the first form can be turned into a simple function of the second form with the same integral, and vice versa, you should be done.

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Does that also hold if $f$ is negative? –  D. Rod Nov 11 '12 at 9:16
    
If we can show it for nonnegative functions, we can extend it to the general case by the linearity of integration. –  Isaac Solomon Nov 11 '12 at 9:18

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