Suppose we have two random variables such that $1 \leq X \leq Y$ almost surely. Now, suppose that all moments of $X$ and $Y$ exist, and,
$$ E[Y^m] \leq E[X^m] + 1 $$
for all $m \geq 1$. Prove that $X$ and $Y$ are equal almost surely.
By assumption, $Y \geq X$ almost surely, and so it suffices to prove that $X \geq Y$ almost surely. Specifically, I would like to show that $P\left(X > Y + \frac{1}{n}\right) = 0$ for all $n\geq1$. My attempt was to manipulate the Markov inequality,
$$ P\left(X > Y + \frac{1}{n}\right) \leq P\left(X > 1 + \frac{1}{n}\right) \leq \dfrac{E[X^m]}{\left(1+\frac{1}{n}\right)^m} \leq \dfrac{C}{\left(1+\frac{1}{n}\right)^m}$$
The first inequality comes from the fact that $Y \geq 1$. The last comes from the fact that $E[X^m] < \infty$ for all $m$. Taking limits as $m \rightarrow \infty$ shows the desired result.
Am I justified in making the last inequality? Specificially, is what I said in bold above true? If the moments were uniformly bounded, then sure, but otherwise I can't be sure that the expression will go to zero.
Notice that $$ Y^m-X^m=(Y-X)(Y^{m-1}+Y^{m-2}X+Y^{m-3}X^2\dots+X^{m-1})\ge (Y-X)m $$ Therefore, $$ E[Y-X]\le \frac1m\cdot E[Y^m-X^m]\le \frac1m\qquad\text{for all }m\ge0 $$ Since $Y-X\ge0$ almost surely, the above implies $Y-X=0$. If you like, you can use the Markov inequality: $$ P(Y-X>\epsilon)\le \frac{E[Y-X]}{\epsilon}\le \frac1{m\epsilon}\stackrel{m\to\infty}\longrightarrow0 $$