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Let $f$ be a bounded measurable function on a set of finite measure $E$. For a measurable subset $A$ of $E$, show that $\int_A f = \int_E f \cdot \chi_A$

Lemma: Let $f$ be a bounded measurable function on a set of finite measure $E$. Suppose $A$ and $B$ are disjoint measurable subsets of $E$. Then $\int_{A \cup B} f = \int_A f + \int_B f$.

My first approach was to view the problem like this utilizing the above lemma:

$$\int_E f \cdot \chi_A = \int_{E\setminus A} f \cdot X_A + \int_A f \cdot \chi_A $$

I'm not really sure how the result follows from the above.

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up vote 1 down vote accepted

$f\chi_A =f$ on $A$, and $f\chi_A = 0$ on $E\setminus A$.

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Does the first part make sense $\int_E f \chi_A$? At first, I wanted to say $\int_E f \chi_E$, but I didn't think that made much sense. –  abet Nov 12 '12 at 6:13
    
The idea is good. $\int_E f\chi_A$ makes sense. –  aid Nov 12 '12 at 6:14
    
I guess, to put my question succinctly, why? My group and I arrived to that by consensus and comfort, but we don't really understand it. –  abet Nov 12 '12 at 6:18
    
(1) The hypotheses of the lemma apply. (2) Applying the lemma how you did and then simplifying gives the desired result. ------- What part is unclear? –  aid Nov 12 '12 at 6:22
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