lebesgue integral - Measure theoretic Fubini/Tonelli theorems: integrating w.r.t variables $t,x,y$ instead of measures $mu, nu, P$?

I've been introduced to the Fubini-Tonelli theorems in my probability/measure theory class in the following way:

or if you want a similar looking pdf: https://www.cmi.ac.in/~prateek/measure_theory/2010-10-06.pdf. Notice that all the integrals in the image and the pdf are w.r.t. measures like $\mu, \nu$ and NOT variables like $t,x,y$.

Question setup: Now, looking at the answer to the following question: Intuitive explanation for $\mathbb{E}X= \int_0^\infty 1-F(x) \, dx$, one of the integrals is $dP$ (a probability measure) and one is $dt$, which is not a measure. This makes me uncomfortable because then it doesn't "look like" we can use the above versions of Fubini/Tonelli on it, because the connotation of $dt$ is Riemann integration and not Lebesgue integration.

Question: can we change the $dt$ integral to be something like $d\lambda$ (where $\lambda$ denotes Lebesgue measure)? When are we sure we can interchange between $dt$ (or $dx$ or $dy$) integrals with integrals w.r.t measures like $\lambda$, especially in these infinite cases?

The reason I specify the infinite case (and the one in the linked question in particular, which is special because the inside function, $f(x) = P(X, is a monotonically decreasing bounded function) is because I know for the bounded, interval case, we have: General condition that Riemann and Lebesgue integrals are the same.

Basically, can someone tell me how to get the $dt,dx,dy$ integrals into the measure theoretic form, and the criterion for knowing when I can do this interchange or not?

I get that this may be a simple misunderstanding of notation on my part, so if that's the case I hope someone can explain the correct way of interpreting this notation.

Answer

All that you need to know are these theorems, that I will state here without proof:

Theorem 1: every Riemann integrable function is Lebesgue integrable, that is, if $\int_a^b f(x)\,\mathrm d x$ exists for some pair $-\infty , then $\int_a^b f(x) \,\mathrm d x=\int_{[a,b]}f\,\mathrm d \lambda $ where the latter is an integral of Lebesgue respect to the Lebesgue measure $\lambda $.

Theorem 2: if $f$ is Lebesgue integrable (w.r.t. the Lebesgue measure in the real line) and the set of discontinuities of $f$ have measure zero then $f$ is improperly Riemann integrable and $\int_{\Bbb R }f\,\mathrm d \lambda =\int_{-\infty}^\infty f(x)\,\mathrm d x$.

Theorem 3: let $f:\Bbb R \to \Bbb R $ Riemann integrable in any bounded interval. Then if $f$ is non-negative or the improper integral of Riemann $\int_a^b|f| \,\mathrm d x$ is finite (for $-\infty\leqslant a\leqslant b\leqslant \infty$) then $\int_{a}^{b}f(x)\,\mathrm d x=\int_{(a,b)}f\,\mathrm d \lambda $, where the first is an improper integral of Riemann, and the latter is an integral of Lebesgue w.r.t. the Lebesgue measure.

Almost any other case of functions $f:I\to \Bbb R $, for some interval $I\subset \Bbb R $, can be reduced to one of the theorems above considering the trivial extension $\tilde f:\Bbb R \to \Bbb R$ defined by

$$
\tilde f(x):=\begin{cases}
f(x),&x\in I\\
0,&\text{ otherwise }
\end{cases}
$$
Hope it helps.