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I was working on a proof of a homework assignment when I stumbled on an issue.

Here is the problem I was working on: Suppose that if $f$ is a nonnegative $\mathcal M$-measurable (Lebesgue measurable) function and that $E\subset \mathcal M$. Let $A$ = {${x\in E}$: $f$($x$) $>$ $0$}. Show that if $\int \limits_{\Bbb E} f d\lambda$ = $0$, then $\lambda (A)$ = $0$.

So far, this is what I have:

Note that $0$ $<$ $f\chi_A$ $\le$ $f\chi_E$. So, we have,

$0$ $\le$ $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ $\le$ $\int \limits_{\Bbb R} f \chi_{E} \,d\lambda$.

By our assumption that $\int \limits_{\Bbb E} f d\lambda$ = $0$, we further see

$0$ $\le$ $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ $\le$ $\int \limits_{\Bbb R} f \chi_{E} \,d\lambda$ = $0$.

This is where I got stuck.
I want to say that $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ = $\lambda$($A$). However, I was told that this does not hold in general.

So my question is: Why does it not hold in general? Under what conditions does it hold?

BCLC
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FoxViking
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3 Answers3

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$\int f\chi_A d\lambda = \lambda (A)$ fails in most cases. Just take A to be an interval and play with some f. If the equation holds for some f do you expect it to hold for $nf$ for all positive integers n? No, right? Here is the correct proof: let $A_n =\{x :f(x)>1/n\}$. Then $\int f\chi_{A_n} \leq \int f\chi_A =0$. On the other hand $f\chi_{A_n} \geq (1/n) \chi_{A_n}$. Hence $\int (1/n) \chi_{A_n} =0$ which menas $\lambda (A_n)=0$. This is true for each n and hence $\lambda (\cup A_n)=0$. now verify that $\cup A_n$ is nothing but A.

Kavi Rama Murthy
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Maybe you already had: For $g\ge 0$ it holds $$\int_{\Bbb R} g \,d\lambda = 0 \iff g \equiv 0 \quad\lambda-a.s$$

Taking $g = f\chi_E$ gives us $$f\chi_E \equiv 0\quad\lambda-a.s$$ from which follows $$f\chi_A \equiv 0\quad\lambda-a.s$$ from what follows $$\chi_A \equiv 0\quad\lambda-a.s$$ what means nothing else then $$\lambda(A) = 0$$

Gono
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  • The crucial step is why $f\chi_A \equiv 0$ $\lambda$-a.s. should imply that $\chi_A \equiv 0$ $\lambda$-a.s. But about this, oddly, you say nothing. – Did Mar 22 '18 at 21:38
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$0$ $\le$ $\int \limits_{\Bbb R} f \chi_{A} \,d\lambda$ $\le$ $\int \limits_{\Bbb R} f \chi_{E} \,d\lambda$ = $0$

$$\iff \int \limits_{\Bbb R} f \chi_{A} \,d\lambda = 0$$

$$\to f \chi_{A} = 0 \ \lambda-\text{a.s.} \tag{1}$$

$$\to \chi_{A} = 0 \ \lambda-\text{a.s.} \tag{2}$$

$$\to \lambda(A) = 0$$

$(1)$ Did you already prove this in class?

If not consider $A_n = \{f(x) > \frac1n \}$

$(2)$ Gasp! What if $f=0$?

BCLC
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  • Not useful. Propagates the OP's mistake, adds to it some empty and unwarranted manipulations of symbols. Yet another use of the site as a personal notepad? – Did Mar 22 '18 at 21:39