Consider a generic probability space $(\Omega,\mathcal{F},\mathbb{P})$ and a measurable space $(\mathbb{R},\mathcal{B})$ where $\mathcal{B}$ is meant to denote the Borel $\sigma$-algebra. A measurable function $$X\,:\,\Omega\rightarrow\mathbb{R}$$ such that $\sigma(X)\subseteq\mathcal{F}$ is then referred to as random variable and allows for the definition of a law (or distribution) $$\mu:\mathcal{B}\rightarrow[0,1]\quad\text{such that}\quad B\mapsto\mathbb{P}(X\in B)\quad\text{for all}\quad B\in\mathcal{B}.$$ Since all intervals of the type $(-\infty,x]$ are elements of $\mathcal{B}$, it follows that one can define a distribution function $$F(x)=\mu(-\infty,x].$$ Due to the properties of $\mu$, $F$ is easily shown to be monotonically increasing, right-continuous, and normalized to take values in $[0,1]$.
Building on these results most texts in probability theory (at least those that I have encountered so far) simply go on asserting that - provided certain conditions on $f$ are met - we can define the Stieltjes integral w.r.t. distribution function $F$ $$\int_a^bf(x)\,\text{d}F(x)$$ to be equal to Lebesgue integral w.r.t. the law of $X$ $$\int_{(a,b]}f\,\text{d}\mu,$$ such that $\int_a^bf(x)\,\text{d}F(x)$ and $\int_{(a,b]}\,f\,\text{d}\mu$ are essentially just two different ways of writing exactly the same thing.
As $F$ and $\mu$ are related by the fact that $F(x)=\mu(-\infty,x]$, I am not really comfortable to accept the above as a matter of definition. Therefore, I was wondering whether there is any way to see that the statement above is actually true?
Thank you very much.
Best wishes,
Jon