Property of function $f=0$ almost everywhere

Question

Theorem: Suppose that $f:X\to [0,\infty]$ is measurable, $E\in \mathfrak{M}$, and $\int \limits_{E}fd\mu=0$. Then $f=0$ a.e. on $E$.

It's very famous fact which can be proven easy.

Am I right that this theorem becomes false if we assume that $f:X\to (0,\infty]$?

I guess that here even condition $\int \limits_{E}fd\mu=0$ fails.

Can anyone exaplain this to me detailed why my assumption false?

score 2 · Accepted Answer · answered May 29 '16 at 09:20

2

Claim: Let $m(E)>0$ and $f : E \to (0,\infty]$. Then $\int_E f>0$.

Proof: Let $E_n = \{ x\in E : f(x) \ge \frac 1n\}$. Then $\cup E_n = E$ and so

$$ \lim_{n\to\infty} m(E_n)= m(E).$$

Since $m(E)>0$, there is $n$ so that $m(E_n) >0$. Thus

$$\int_E f \ge \int_{E_n} f \ge \int_{E_n} \frac 1n = \frac 1n m(E_n) >0.$$

answered May 29 '16 at 09:20

So my claim is false right? Because integral can be positive. – RFZ May 29 '16 at 09:22
So your guess is correct, there isnt such a function. @RFZ – May 29 '16 at 09:23
You showed that integral is positive. Why there isnt such a function? – RFZ May 29 '16 at 09:25
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You mean a function with zero integral right? – RFZ May 29 '16 at 09:26
@RFZ Yes. ${}{}{}{}$ – May 29 '16 at 09:27
Dear John! Can you check my proof please? http://math.stackexchange.com/questions/1810581/every-separable-metric-space-has-a-countable-base – RFZ Jun 03 '16 at 07:20

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