How is the "Push-Forward Measure" a measure?

Question

I just watched a video on Measure Theory and am quite sure I had some misunderstanding as to how a measurable map works. The video asserted that a push-forward measure is a proper measure, so for some reason or another, the following counter-example must be wrong:

$$S = \sigma = \{ \{\}, \{a\}, \{b\}, \{a,b\} \}$$

$$ \mu(\{\})= 0, \mu(\{a\})= \mu(\{b\})=1, \mu(\{a,b\})=2 $$

$$f(S,\sigma) \rightarrow (S,\sigma) := f(\{\})=\{\}, f(\{a\})=\{a\}, f(\{b\})=\{a,b\}, f(\{a,b\})=\{b\}$$

Wouldn't then the resultant push-forward measure not work as $f_* \mu(\{a\})+f_* \mu(\{b\}) \neq f_*\mu(\{a,b\})$? I realize I must have something misunderstood, so some clarification would be much appreciated.

Answer 1

The function $f$ has to be defined on the underlying sets ($S$) of the measurable spaces, not their $\sigma$-algebras $(\sigma$). So your supposed example does not apply.

Answer 2

The problem is the following.

Let $S$ be a set and let $\mathcal{A}$ be a $\sigma$-algebra on $S$, that is a collection of subsets of $S$ satisfying certain properties. The couple $(S,\mathcal{A})$ is referred to as a measurable space. A measurable function $f$ between two measurable spaces $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ is a function $f:X\rightarrow Y$ such that $f^{-1}(B)\in \mathcal{A}$ for all $B\in \mathcal{B}$. Thus, to speak of "measurable functions" you need functions defined over spaces, and not over their $\sigma$-algebras; in your example you need a funxction $f:S\rightarrow S$.