I was asked to show that $g_f(x)=\mu(f\leq x)$ defines a cumulative distribution function for any measurable function $f$.
Let $(\Omega,\mathcal{F},\mu)$ be a measure space and $(\mathbb{R},\mathcal{B})$ a measurable space. As far as I understand, $\mu(f\leq x)=\mu(\{w\in\Omega:f(w)\leq x\})=\mu(f^{-1}(-\infty, x])$ for $f:\Omega \rightarrow \mathbb{R}$. Given these measurable spaces, $f$ induces the pushforward measure on $\mathcal{B}$ defined by $\mu_f((-\infty, x])=\mu(f^{-1}(-\infty, x])$. But to make $F_{\mu_f}(x)=\mu_f((-\infty, x])$ to be the cumulative distribution function of $\mu_f$, I think it is needed that $\mu$ be finite.
Is my argument going in the right direction?
I'm feeling that something is missing there. Could someone organize the concept of CDF associated to a measure for a better understanding?
Thanks in advance!
The definition I hve for a cumulative distribuition function is as follows:
Let $I=[a,b]\subseteq \mathbb{R}$, $\mathcal{B}(I)$ be the Borel $\sigma$-algebra and let $\mu$ be a finite measure on $(I,\mathcal{B}(I))$. A cumulative distribution function of $\mu$ is a function $g_{\mu}:I\rightarrow\mathbb{R}$ defined by $g_{\mu}(x)=\mu[a,x],\forall x \in I$.
The definition above seems to require that the original statement be restricted to
Show that $g_f(x)=\mu(f\leq x)$ defines a cumulative distribution function for any measurable function $f:(\Omega,\mathcal{F})\rightarrow (I,\mathcal{B}(I)) $ and any $\mu$ finite on $\mathcal{F}$.
So if I assume $f:\Omega\rightarrow I, (\mathcal{F},\mathcal{B}(I))$-measurable and assume $\mu$ finite on $(\Omega,\mathcal{F})$, then $\mu_f=\mu\circ f^{-1}$ will be finite on $(I,\mathcal{B}(I))$ and I might finish the proof, on the lines of my argument above.