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There's an exercise in my book that states the following:

Let $\mu_0 (A) = \mu(A \cup A_0),\,A \in \mathcal{F}$ be a measure. Show that if $$\int f \,d\mu$$ exists, then $$\int_{A_0} f\,d\mu = \int f\,d_{\mu_0}$$

However, I think there's a typo in this exercise. $\mu_0$ isn't even a measure when $\mu(A_0) \neq 0$, since $\mu_0(\varnothing) = \mu(A_0)$. If it were $\mu_0(A) = \mu(A \cap A_0)$, which I think that's what the author meant, I could prove it in the following way:

For simple functions,

$$\int_{A_0}f\,d\mu = \int f\mathbf{1}_{A_0}\,d\mu = \sum_{i=1}^{n} f_i \mathbf{1}_{A_0}\mu(A_i) =\sum_{i=1}^{n} f_i\mu(A_i \cap A_0) = \sum_{i=1}^{n} f_i \mu_0(A_i) = \int f\,d\mu_0$$

For a non-negative measurable function $f$, there's a non-decreasing sequence of simple functions $(s_n)_{n \in \mathbb{N}}$ such that $s_n \rightarrow f$, then

$$\int_{A_0} f\,d\mu= \int f\mathbf{1}_{A_0}\,d\mu = \lim_{n \rightarrow \infty} \int s_n\mathbf{1}_{A_0}\,d\mu = \lim_{n \rightarrow \infty} \int s_n\,d\mu_0 = \int f\,d\mu_0$$

And finally, for any measurable function, we just use $f = f^{+} - f^{-}$. So, which one is it, $\mu(A \cup A_0)$ or $\mu(A \cap A_0)$?

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  • $\begingroup$ The first thing you show is that $\mu(A_0) = 0$ in order that $\mu$ is a measure. So $A_0$ has zero $\mu$-measure... (In fact, that's pretty much the only thing the hypotheses tell you about $A_0$.) $\endgroup$ Nov 25 at 2:11
  • $\begingroup$ @EricTowers Yes, but if $\mu(A_0) = 0$ then $\int_{A_0} f\,d\mu$ would also have to be zero. And, because the author defined $\mu_0 = \mu(A_0 \cup A)$, I find it highly unlikely that $\int f\,d\mu_{0} = 0$. $\endgroup$ Nov 25 at 2:15
  • $\begingroup$ So, ..., what do you think the $\mu_0$-measure of the support (or a level set) of $f$ is? $\endgroup$ Nov 25 at 2:17
  • $\begingroup$ @EricTowers I'm sorry, I don't know what you mean. I asked whether the definition of $\mu_0$ given by the author was wrong. $\endgroup$ Nov 25 at 3:37
  • $\begingroup$ It is not wrong. And, no you didn't. $\endgroup$ Nov 25 at 3:37
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Your proof looks good (with some minor typo). I think you are right that it should be $\mu_0(A) = \mu(A \cap A_0)$.

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  • $\begingroup$ I was thinking about some counterexamples. If we assume $\mu_0(A)=\mu(A \cup A_0)$ then $\int_{A_0} \mathbf{1}\,d\mu = 0$ (because we need to have $\mu(A_0) = 0$ given that $\mu_0$ is a measure), but $\int \mathbf{1}\,d\mu_0 = \mu_0(\Omega) = \mu(\Omega \cup A_0) = \mu(\Omega) \neq 0$. So, in this case, $\int_{A_0} \mathbf{1}\,d\mu \neq \int \mathbf{1} d\mu_0$. $\endgroup$ 2 days ago

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