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I'm trying to show that given two regular Borel measures on a compact space $X \subset \mathbb{R}$, $\mu_1, \mu_2$, and a bounded Borel function $f: X \to \mathbb{C}$, there exists a sequence $\{f_n\} \subset C(X)$ of continuous functions s.t $\int_X f_n d\mu_1 \underset{n \to \infty}{\to} \int_X f d\mu_1$ and $\int_X f_n d\mu_2 \underset{n \to \infty}{\to} \int_X f d\mu_2$.

My attempt:

Define $\mu = \mu_1 + \mu_2$, so that $\mu$ is regular Borel measure.

We know that there exists $\{f_n\} \subset C(X)$ s.t $\mu(\{x \in X : lim_n f_n(x) \neq f(x)\}) = 0$.

It follows that $f_n \underset{point-wise}{\to} f$ a.e$[\mu_i]$ for $i = 1,2$.

If I can choose $f_n$ so that $\{||f_n||_\infty\}$ is bounded, my question will follow from the dominated convergence theorem.

Can such a sequence be chosen, despite that we only have a.e point-wise convegence to the bounded function $f$? If so, is the rest of the argument ok? If not is there another method to show this?

Mariah
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    Perhaps I am being naive here, but wouldn't it be enough to take a sequence $f_n$ of continuous functions which approximate $f$ in the measure $\mu = |\mu_1| + |\mu_2|$? Because $\int_X |f_n-f| d\mu_1 \leq \int_X |f_n-f| d\mu$. – Bartosz Malman Oct 20 '17 at 18:27
  • @Malman mm nice idea, wondering whether we're missing something – Mariah Oct 20 '17 at 18:49
  • Since $f$ is bounded, doesn't $|f_n|_\infty<\infty$ follow, possibly after adjusting the sequence of these continuous functions? Afterall, $X$ is compact. – Alex R. Oct 20 '17 at 19:32
  • By the crollary of Lusin's theorem, you can choose a sequense ${f_n}$ of continuous functions such that $f_n\to f $ pointwise and $\lVert f_n\rVert_\infty\leq \lVert f\rVert_\infty$. I thought about something like $\mu_1+\mu_2$ but gave up at that time. Now I think you are right, and congratulate you find an easier way. – C. Ding Oct 21 '17 at 14:38
  • @C.Ding you say, given a bounded Borel function $f$ there is a sequence of continuous, uniformly bounded function which converge to $f$ point-wise. If that were true it would follow by the dominated convergence theorem that for any $\mu$, a regular Borel function, $\int f_n d\mu \to \int f d\mu$, which I thought we contradicted. – Mariah Oct 21 '17 at 14:44
  • Not for any $\mu$, just for one measure by Lusin's theorem and for finite measures using your argument. Or even for countable measures using $\sum_{n=1}^{\infty} \frac{\mu_n}{2^n\mu_n(X)}$? – C. Ding Oct 21 '17 at 14:54
  • @C.Ding ah ok, so you mean a.e point wise convergence I guess? btw, I'd really appreciate your input in my question of commuting operators, which is linked to this thread as well. – Mariah Oct 21 '17 at 14:59

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