this might seem a stupid question but I'm struggling to answer it myself.
From beginner's probability we know that the standard normal distribution has density $$\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$$
However, I am reading Le Gall's Brownian Motion, Martingales and Stochastic Calculus where he defines the former as a density with respect to the Lebesgue measure.
So this got me thinking, is it trivially obvious that
$$\int_{-\infty}^t\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}dx=\int_{(-\infty,t)}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}d\mu$$
where $\mu$ is the Lebesgue measure?
Obviously, the chosen density has no specific properties(except possibly that it integrates to $1$) but this is not the point of the question.
I know (though the proof was not given to me) that $\mu((a,b))=b-a$ but still, can't we have a cover of $(-\infty, t)$ in my specific case on top which consists of more complicated elements?
Hopefully the question makes sense.
This Criterion for an Improper Riemann Integrable Function to be Lebesgue Integrable should help. One needs to check that the Riemann integral of the absolute value of the improper integrable function is bounded by the same positive constant over all finite intervals.