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I'm struggling to prove the following Lemma from V.I. Bogachev, Measure Theory 2:

Let two $\tau$-additive measures $\mu$ and $\nu$ on a topological space $X$ coincide on all sets from some class $\mathcal{U}$ that contains a base $\mathcal{B}$ of the topology $\mathcal{T}$ in $X$ and is closed with respect to finite intersections. Then $\mu=\nu$.

Recall that a Borel measure $\mu$ is called $\tau$-additive on a topological space $X$ if for every increasing net of open sets $(O_i)_{i\in I}$ in $X$ one has the equality $$|\mu|\bigg(\bigcup_{i\in I}O_i\bigg)=\lim_{i\in I}|\mu|(O_i)$$

The given proof is a little minimalistic:

Every open set $O\in\mathcal{T}$ can be represented in the form of the union of a net of increasing open sets $O_i$ that are finite unions of sets in $\mathcal{U}$. It is easily seen that $\mu(O_i)=\nu(O_i)$ for all $i\in I$. By the $\tau$-additivity we obtain $\mu(O)=\nu(O)$. Since the two measures coincide an all open sets, they coincide on all Borel sets.

With some help from Daniel Fischer I've constructed the net of increasing open sets $(O_i)_{i\in I}$ via the directed set $(I,\subseteq)$, where $$I:=\{\mathcal{B}_i\subseteq\mathcal{B}\,|\,\mathcal{B}_i\text{ is finite and }\forall B\in\mathcal{B}_i\,:\,B\subseteq O\}. $$ and $$O_i:=\bigcup_{B\in\mathcal{B}_i}B $$ This yields $O_i\subseteq O_j$ whenever $B_i\subseteq B_j$, $\bigcup_{\mathcal{B}_i\in I}O_i=O$ and every $O_i$ is a finite union of elements of the base $\mathcal{B}$.

Now, I have troubles showing $\mu(O_i)=\nu(O_i)$ for all $\mathcal{B}_i\in I$. Also, I don't see where the assumption that $\mathcal{U}$ is closed with respect to finite intersections is needed.

Any help would be highly appreciated and thank you very much in advance for your efforts!

Greetings

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Use inclusion-exclusion: For any finite set $F$ we have \begin{eqnarray*} \mu\left(\cup_{i\in F} B_i\right) &=&\sum_{\emptyset \subset J\subseteq F}(-1)^{|J|+1}\mu\left(\cap_{j\in J} B_j\right)\\[5pt] &=&\sum_{\emptyset \subset J\subseteq F}(-1)^{|J|+1}\nu\left(\cap_{j\in J} B_j\right)\\[5pt] &=&\nu\left(\cup_{i\in F} B_i\right), \end{eqnarray*} where the sum runs over all non-empty subsets $J$ of $F$. For the second equation you use the fact that $\cal B$ is closed under finite intersections.

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  • $\begingroup$ I'd never thought of that, thank you very much! $\endgroup$ – Florian Jul 25 '15 at 15:35

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