# Question on Proof: when two measures agree on $\pi$-system, then they agree on generated $\sigma$-algbra

Let $\mu_1, \mu_2$ be measures on $\sigma(\mathcal P)$, where $\mathcal P$ is a $\pi$-system, and suppose they are $\sigma$-finite on $\mathcal P$. If $\mu_1$ and $\mu_2$ agree on $\mathcal P$, then they agree on $\sigma(\mathcal P)$.

My proof: Define $$\mathcal G := \{ A \in \sigma(\mathcal P) : \mu_1(A) = \mu_2(A) \}.$$ Then $\mathcal P \subseteq \mathcal G$ because they agree on $\mathcal P$. Further $\mathcal G$ is a $\sigma$-algebra:

i) $\Omega \in \mathcal G$.

Let $\Omega = \bigcup_k P_k$ where the $P_k$ are disjoint, and $\mu_1(P_k) < \infty$ and $P_k \in \mathcal P$ (and so $\mu_2(P_k) = \mu_1(P_k)$), such a decomposition exists because of $\sigma$-finiteness. Then $$\mu_1(\Omega) = \mu_1\left( \bigcup_k P_k \right) = \sum_k \mu_1(P_k) = \sum_k \mu_2(P_k) = \mu_2\left(\bigcup P_k\right) = \mu_2(\Omega)$$ and so $\Omega \in \mathcal G$.

ii) Closure under complement.

Let $A \in \mathcal G$, then $A \in \sigma(\mathcal P)$ and $\mu_1(A) = \mu_2(A)$, by this $$\mu_1(A^c) = 1 - \mu_1(A) = 1 - \mu_2(A) = \mu_2(A^2)$$ and so $A^C \in \mathcal G$.

iii) Closure under countable union.

Let $\{ A_n \} \subseteq \mathcal G$. Then $\bigcup A_n \in \sigma(\mathcal P)$ because it is a $\sigma$-algebra and $$\mu_1\left( \bigcup_n A_n \right) = \sum_n \mu_1( A_n ) = \sum_n \mu_2( A_n ) = \mu_2\left( \bigcup_n A_n \right)$$ and so $\bigcup A_n \in \mathcal G$.

With $\mathcal P \subseteq \mathcal G$ we have $\sigma(\mathcal P) \subseteq \mathcal G$, and more specific $\sigma(\mathcal P) = \mathcal G$, so that both measures agree on all sets of $\sigma(\mathcal P)$. $\square$

Is this proof okay? I came up with it myself, but when I look it up in a book, the one I found is much more complicated, which makes me suspicious, because my proof is quite simple and straight, but I do not see where it could fail?

• I am not totally convinced about i. About ii) you use $\mu_1(\Omega) = \mu_2(\Omega) = 1$. Indeed i knew this theorem in case of finite measures with same total measure. This makes me also more confident on 1. On iii) you showed the assertion only for sequence of pairwise disjoint events. – Kolmo Jan 7 '15 at 22:09
• Okay, ii) and iii) are not enough. Regarding i) let me add why the decomposition $\Omega = \bigcup_k P_k$ could be choosen disjoint. If they are not, replace $P_k$ by $P_k \cap P_1^C \cap \ldots \cap P_{k-1}^C$. – StefanH Jan 7 '15 at 22:20
• It could be that (for example) $\mathcal{P} = \{M \subset X \mid x_0 \in M\}$ with some fixed $x_0$. Then you can not "disjointify" while staying in $\mathcal{P}$. As noted in the other comment, (iii) only works for disjoint unions. The usual proof for this uses Dynkin's $\pi-\lambda$-Theorem to circumvent this problem. (See en.wikipedia.org/wiki/…) – PhoemueX Jan 7 '15 at 22:27