Restriction of a measure to a sigma-algebra

398 Views Asked by At

Let $X$ be a set and $A,B$ two nonempty proper subsets of $X$ and $A \cup B \neq X$.

Define $\varepsilon := \lbrace A,B \rbrace \in \mathcal{P}(X)$.

$\mathcal{\sigma}(\varepsilon)$ is the sigma-algebra generated by $\varepsilon$ and $\lambda : \varepsilon \to (0,\infty)$ a map.

How to prove that two different measures $\mu, \mu':\mathcal{\sigma(\varepsilon) \to [0,\infty]}$ exist with $\mu' \big|_{\varepsilon}=\mu \big|_{\varepsilon}=\lambda$ ?

I thought about showing that they give the same measure to each subinterval, but I don't know how to show the uniqueness on the restriction on $\varepsilon$.