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Note: This wall of text is overkill and probably not too representative of the question in the title. I recommend you simply skip to edit 4, and then see the answers below.

Let us consider the integral $\int_A fd\mu = \int 1_A(x)f(x)d\mu (x), A\in \mathscr{B}$ (we are dealing with Borel sets).

If $\mu (A) = 0$ then the integral is zero, and thus the integrand is zero (as referenced by many other topics here on stack exchange, such as This topic).

What I don't get is how $\mu(A) = 0$ implies the integrand is zero. I believe it implies that the integral is zero, as $\int 1_A (x) f(x) d\mu (x)$ just $f \mu (A)$ = 0$, correct? But I don't see how the integrand must be zero.

Indeed, even at the thread I linked to, it seems that it suffices to show that if $\mu =0$, or $\int_A fd\mu =0 $then $f=0$, but I don't understand how those lead to $f=0$...

Edit: maybe it is something like $f = 1_A (x) f(x)$ which can be anything where $\mu =0$, and in order for $\int 1_A(x)f(x)d\mu (x) =0 $ (implied by $\int_A f d\mu = 0$ and $\int_A f d\mu = \int 1_A(x)f(x)d\mu (x)$) then $1_A(x)f(x) = 0$ a.e., and perhaps $1_A(x)f(x) = f(x)$ somehow?

Edit 2: I guess here is what I am wondering: in the problem $\int_A fd\mu$, is the integrand $f$ or is it $f$ on the domain A. For those who don't follow, here is why it matters: if the integrand is $f$ on the domain $A$, then we know that $f$ on the domain $A$ is $1_A(x) f(x)$. But $A$ has measure zero since $\mu(A) = 0$, and $1_A(x)f(x) = 0$ everywhere outside A (because the indicator function is zero), so therefore we can say that the integrand $f$ is zero A.E (since A.E. means we exclude (don't consider) set A since it has measure zero)

EDIT 3: I will ask the above edit (#2) as a separate question and link the answer here, hopefully.

EDIT 4: (Hopefully final) So I believe I asked a question different than what I had in mind (but which worked out better for me), and perhaps different than what I outlined in the mass of text above. here is what the title asks: If $\mu (A) =0$, and $\int f d\mu = 0$, then $f=0$ a.e. The reasoning for this is as both people state below, albeit briefly. If the measure of a set that is not A is not zero, then the integrand $f$ must be zero (in order for the integral to be zero). Whatever $f$ is on A, however, we don't care about, since we are saying that the function is zero A.E.

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majmun
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2 Answers2

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The implication is only that the integrand is zero almost everywhere, i.e., everywhere but on a set of $\mu$ measure zero. Thus, you do not get any information about the values of $f$ on $A$ because $A$ has measure zero.

According to your edit 2 there are two viewpoints, either you consider the function $f$ as your integrand on the domain $A$ or you consider the function $f 1_A$ as your integrand on the whole space. In either case you can infer that the integrand is $0$ almost everywhere. In the first case this implies that $f$ is zero almost everywhere on $A$ (which has no meaning because $A$ has measure zero) and in the second case $f 1_A$ is zero almost everywhere on the whole space which is again obvious because $1_A$ is zero on $A^c$ and again you don't care for the values on $A$.

KoliG
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  • +1. Re no meaning: one can consider this means that ${f\ne0}$ has measure zero for the measure $\mu$ restricted to $A$ (which, as you explain, is an empty condition since this restricted measure is itself null). – Did Nov 13 '15 at 13:09
  • I added a few more edits, but basically I was thinking of a question other than what the title was describing. Sorry for the confusion -- your original answer was enough. – majmun Nov 13 '15 at 13:09
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Integral is zero iff integrand is zero a.e. If $\mu(A) = 0$ then $\mu(1_A\cdot f \neq 0) = 0$ for any $f$.

SBF
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