# Uniqueness of measures

Let $\Omega \neq \emptyset$ and $\mathcal{F}$ a $\sigma$-algebra of subsets of $\Omega$.

Let $\mu_1, \mu_2: \mathcal{F} \to [0, \infty)$ be measures.

Theorem. Let $\mathcal{C} \subset \mathcal{F}$ be a $\pi$-system such that $\mathcal{F} = \sigma\langle \mathcal{C}\rangle$. If $\mu_1(C) = \mu_2(C)$ for all $C \in \mathcal{C}$ and $\mu_1(\Omega) = \mu_2(\Omega)$, then $\mu_1(A) = \mu_2(A)$ for all $A \in \mathcal{F}$.

I wish to show that $\mathcal{L}:= \{A: A \in \mathcal{F}, \mu_1(A) = \mu_2(A)\}$ is a $\lambda$-system. There are three parts, as far as I know:

1. $\Omega \in \mathcal{L}$ trivially, since $\mu_1(\Omega) = \mu_2(\Omega)$ by assumption and $\Omega \in \mathcal{F}$.
2. Let $A, B \in \mathcal{L}$ satisfy $A \subset B$. I need to show that $B \setminus A \in \mathcal{L}$. $B \cap A^c \in \mathcal{F}$ since $\mathcal{F}$ is a $\sigma$-algebra (already proved closure under finite intersections for a $\sigma$-algebra). Next, I have to show that $\mu_1(B \setminus A) = \mu_2(B \setminus A)$. Observe \begin{align} &\mu_1(B \setminus A) = \mu_1(B \cap A) + \mu_1(B) \\ &\mu_2(B \setminus A)= \mu_2(B \cap A) + \mu_2(B) \end{align} by a theorem I previously proved. We have $\mu_1(B) = \mu_2(B)$ by definition of $\mathcal{L}$. But I'm not sure how to show $\mu_1(B \cap A) = \mu_2(B \cap A)$; do we know if $B \cap A \in \mathcal{C}$?
3. I have to somehow construct $\{A_i\}_{i=1}^{\infty} \subset \mathcal{L}$ such that $A_n \subset A_{n+1}$ for all $n \geq 1$ implies that $\bigcup_{n=1}^{\infty}A_n \in \mathcal{L}$. But I don't even know where to start with this.

## 1 Answer

1. $\mu_1$ and $\mu_2$ are finite measures, so you can write $$\mu_1(B\setminus A)=\mu_1(B)-\mu_1(A)$$ $$\mu_2(B\setminus A)=\mu_2(B)-\mu_2(A)$$ Hence the result.

3.Let $(A_i)_{i\geq 1}$ an increasing sequence of elements of $\mathcal{L}$. Then $\bigcup_{i=1}^{\infty}A_i\in\mathcal{F}$ and by sequential continuity of measure, $$\mu_1\left(\bigcup_{i=1}^{\infty}A_i\right)=\lim\limits_{i\to\infty}\mu_1(A_i)=\lim\limits_{i\to\infty}\mu_2(A_i)=\mu_2\left(\bigcup_{i=1}^{\infty}A_i\right)$$ So $\bigcup_{i=1}^{\infty}A_i\in\mathcal{L}$.

• Never mind, that was a stupid question. :) – Clarinetist Feb 29 '16 at 14:21