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I am having trouble with a certain technical part of solving a problem. The question can be formulated as the following:

Let $X$ be a locally compact Hausdorff space and $M(X)$ be the space of all regular Borel complex measures on $X$. Pick any $\mu, \nu \in M(X)$. We have that

$$\int f \ d\mu = \int f \ d\nu$$ for every bounded continuous function $f$ on $X$.

Is it sufficient to conclude that $\mu = \nu$? If not,then what additional condition is needed?

I believe that since $C_0(X) \subset C_b(X)$ and $C_0(X)$ is a separating family of functions, it should be true.

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    $\begingroup$ The Riesz Representation Theorem might help $\endgroup$
    – Araske
    Dec 5, 2017 at 4:46
  • $\begingroup$ I have edited the answer. have a look. $\endgroup$
    – Mayuresh L
    Dec 7, 2017 at 7:36

1 Answer 1

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First of all as complex measures are finite, for any $f\in C_b(X)$ and $\mu \in M(X)$

$\int f \ d\mu$ exists

Let $m=\mu-\nu$ hence $\int f \ dm=0$ for any $f\in C_b(X)$

Note $m$ is a regular complex measure. Now there exist complex valued function $h$ s.t $|h|$=1 a.e. and

$m(A)= \int_A h d|m|$

Now $m(X)= \int_X h d|m|= \int_X (\bar h-f)h d|m| \leq \int_X|\bar h-f| d|m| \leq \epsilon$

for sutaible choice of $f\in C_b(X)$ (because $C_b(X)$ is dense in $L^1(|m|))$

which gives $|m|(X)=0$

i.e. $m=0$

i.e. $\mu = \nu$.

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