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I've been trying to prove the following statement: Suppose a measure space $(X, M, \mu)$. If $\int_E f \: d\mu = 0$ for every measurable set $E \in M$, then $f = 0$ almost everywhere in $X$. I've definitively proved it for the case when $\mu$ is a positive measure, but the complex case completely eludes me. I've already proven that if $f = u + iv$, where $u$ and $v$ are real functions, then $u = 0$ almost everywhere and $v = 0$ almost everywhere. To me it seems really clear then that $f = 0$ almost everywhere, but the formal definitions we are working with is the following:

A property P holds almost everywhere in a set E if there is a measurable set N such that P holds in E \ N, and the measure of N is zero.

I therefore know that there are sets $A_u$ and $A_v$ with measure zero such that $u = 0$ in $X \setminus A_u$, and analogous for $A_v$. What I don't know yet is if $A_u \cup A_v $ has measure zero.

Or maybe I'm going through this the wrong way.

Sebastian Garrido
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  • You might check this question out. – RCT Mar 14 '17 at 04:54
  • Can't you just take the union of the null set where $u\neq 0$ with the null set where $v\neq 0$? The union of finitely many measure zero sets has measure zero – Nick Alger Mar 14 '17 at 04:58
  • The OP specifically mentions that the measure is not assumed to be positive. – RCT Mar 14 '17 at 05:07
  • @RCT I think the question you referenced assumes a positive measure, though I get the feeling that the proof there is somewhat salvageable for the complex case. I'll think about it for a while. – Sebastian Garrido Mar 14 '17 at 05:29
  • @SebastianGarrido I was suggesting that it was adaptable, not immediately applicable, but I see now that the thing I was thinking of makes the same mistake I pointed out above. Whoops! – RCT Mar 14 '17 at 05:49

1 Answers1

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If $\mu$ is a complex measure, then there exists a measurable function $h$ with $|h(x)| = 1$ for all $x \in X$ such that $$ d \mu = h \ d|\mu|.$$ Here, $|\mu |$ is the total variation measure associated to $\mu$. This statement follows from the Radon-Nikodym theorem. (See Rudin 6.12.)

Since $\int_E f d \mu = 0$ for all measurable $E$, it follows that $$ \int_E f(x) h(x) d |\mu| = 0$$ for all measurable $E$.

But the total variation measure $|\mu |$ is a positive measure! From the discussion in the comments, we know that your result applies to positive measures! So we deduce that $$ f(x) h(x) = 0$$ almost everywhere with respect to $|\mu|$.

Finally, $|h(x)| = 1$ for all $x \in X$. Hence $f(x) = 0$ almost everywhere with respect to $|\mu |$. Since any set that is null with respect to $|\mu |$ also has zero measure with respect to $\mu$, your result follows.

Kenny Wong
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    Thanks! We haven't seen the Radon-Nikodym theorem, but I'll catch up on it. – Sebastian Garrido Mar 16 '17 at 13:14
  • No worries! I must admit I'm not a measure theory expert and I've never studied the proof of Radon-Nikodym. But I find it helpful to have memorised some of its consequences. This fact that any complex measure is equal to a function of unit modulus times its total variation measure is something I find really useful. One day I hope to study that proof... – Kenny Wong Mar 16 '17 at 16:16