Proving the Probability Integral Transformation Theorem using MGF 
Prove the probability-integral transformation, i.e., if $F_X$ is
continuous, then $F_X(x)\overset{d}{=}\mathsf{Unif}(0,1)$, by finding
the mgf of the random variable $Y=F_X(X)$ where $X$ is absolutely
continuous and has cdf $F_X$.

This is easy to show by noting that
$$\mathsf P\left(F_X(X)\geq x\right)= \mathsf P\left(X\geq F_X^{-1}(x)\right) = 1-F_X\left(F_X^{-1}(x)\right)=1-x\tag{1}$$
but I'm having trouble showing this by mgf. Since the mgf of a $\mathsf{Unif}(0,1)$ random variable is given by $\frac{e^t-1}{t}$ then we need to show that
$$M_Y(t)=\int_{-\infty}^{\infty} e^{tY} f_Y(y)dy = \frac{e^t-1}{t}$$
The only way I can think about showing this is by noting that
$$\begin{align*}
M_Y(t)
&=\int_{-\infty}^{\infty} e^{ty} f_Y(y)dy\\\\
&= \int_{-\infty}^{\infty} e^{ty} dF_Y(y)\\\\
&= \int_{0}^{1} e^{ty} dy\\\\
&=\frac{e^t-1}{t}
\end{align*}$$
but this requires the knowledge that $F_Y(y)=y$ which is already sufficient to showing that $Y\sim\mathsf{Unif}(0,1)$. Is there a way to show this without making use of (1)?
 A: Your idea is the correct one, but you need to take a slightly different route to get to the end.
To be rigorous, we ought to be working with the characteristic function (cf) of $Y=F_X(X),$ which (unlike the mgf) always exists and determines the distribution.  (If you don't like this, simply erase all occurrences of "$i$" in the following.)
Let's just plug stuff into the definition of the cf, which begins these equalities.
$$\begin{aligned}
E\left[\exp(it\,Y)\right] &= E\left[\exp\left(it\,F_X(X)\right)\right] & \text{definition of } Y\\
&= \int_{\mathbb{R}} \exp\left(it\, F_X(x)\right)\,\mathrm{d}F_X(x)& \text{expectation formula} \\
&= \int_0^1 \exp(it\,y)\,\mathrm{d}y & \text{formal substitution }y=F_X(x)\\
&= \frac{e^{it}-1}{it}.
\end{aligned}$$
This derivation only required knowing that

*

*$F_X$ is the cdf of a continuous variable (if you use the Lebesgue-Stieltjes integral) or at least differentiable (if you use the Lebesgue integral) and


*$F_X:\mathbb{R}\to[0,1]$ is a one-to-one function.
No assumptions about $F_Y$ were made (or harmed) in the derivation of this result.
