Counter-example: measures that agree on a collection of subsets but not on the generated sigma-algebra

Question

From Bass, Real Analysis for Graduate Students:

Suppose $(X,\mathcal A)$ is a measurable space and $\mathcal C$ is an arbitrary subset of $\mathcal A$. Suppose $m$ and $n$ are two $\sigma$-finite measures on $(X,\mathcal A)$ such that $m(A)=n(A)$ for all $A\in \mathcal C$. Is it true that $m(A)=n(A)$ for all $A\in \sigma(\mathcal C) ?$ What if $m$ and $n$ are finite ?

I'm persuaded both are not true since $\mathcal C$ need not be a $\pi$-system (that is to say closed under intersection).

I've been looking for a counter-example, but the finiteness assumptions make it difficult. For example, on $(\mathbb R, \mathcal B(\mathbb R))$, let $\mu(\emptyset)=0$ and $\mu(A)=\infty$ whenever $A$ is a nonempty Borel set. $\mu$ and the counting measure agree on all open sets, but obviously not on $\mathcal B(\mathbb R)$. Nevertheless $\mu$ is not $\sigma$-finite.

Can someone give some hints ? My intuition is still lacking.

Answer 1

This is my previous comment:

Take $X=\{1,2\}$, $\mathcal{C}=\{\{1\}\}$ and measure $m,n$ on $\sigma(\mathcal{C})$ with $m\{1\}=m\{2\}=n\{1\}=1$ and $n\{2\}=2$. Note that in this case $\mathcal{C}$ is a $\pi$-system.

The problem with $\mathcal{C}$, which allows us to construct counter-examples, is that we do not necessarily have $m(X)=n(X)$.

If we assume that $m(X)=n(X)$ and $m$ and $n$ are finite, then the finite conclusion holds: $m(A)=n(A)$ for all $A\in\sigma(\mathcal{C}))$ (to prove this, simply show that the family $\left\{A:m(A)=n(A)\right\}$ is a $\sigma$-algebra containing $\mathcal{C}$).

Answer 2

Let $L$ be Lebesgue measure on the reals. For $L$-measurable $A$ let $$m(A)=2L(A \backslash [0,1])+L(A\cap [0,1]),$$ $$n(A)=L(A \backslash [0,1])+2L(A\cap [0,1]).$$ Let $C=\{c_1,c_2\},$ where $c_1=[-1,1]$ and $c_2=[0,2].$ Then $m(c_1)=m(c_2)=n(c_1)=n(c_2)=3,$ but $m(c_1\cap c_2)=1$ and $n(c_1\cap c_2)=2.$

If we want finite $m,n$ we can restrict the space to $[-1,2]$ rather than $\mathbb R$.