$\boxed{\text{Let } X,Y \text{ be any random variables. Prove that if } \mathbb{E}(|X-Y|) = 0, \text{then } X^2 = Y^2.}$
We have $\mathbb{E}(|X-Y|) = 0$, and I thought that if I could somehow show $X=Y$ then it directly implies $X^2 = Y^2$, but I'm not sure how to get there.
P.s.: This is not a homework assignment of any kind, I just saw this question in one of my university's Probability course past exams and wanted to try it out.
The asserted result is false, but a slightly weaker result is true (see theorem and proof below). The weaker result is simply that you have to add "almost surely" to the asserted equivalence. This is because it is possible to construct random variables that can take on arbitrary values that do not affect their mean. It is also worth noting that the antecedent condition here also implies that $X=Y$ almost surely, which is a stronger result than the one I have shown below (proof of the latter is very similar).
Lemma: If $\mathbb{P}(|Z|> \epsilon)=0$ for all $\epsilon>0$ then $\mathbb{P}(Z=0)=1$.
Proof: Suppose we take any sequence of positive decreasing values $\epsilon_1 > \epsilon_2 > \epsilon_3 \cdots$ with $\lim_{n \rightarrow \infty} \epsilon_n = 0$ and let $\mathscr{Z}_n \equiv \{ |Z| \leqslant \epsilon_n \}$. We may observe that: $$\limsup_{n \rightarrow \infty} \mathscr{Z}_n = \bigcap_{n=1}^\infty \bigcup_{m=n}^\infty \mathscr{Z}_m = \{ Z=0 \}.$$ For all $n \in \mathbb{N}$ we have $\mathbb{P} (\mathscr{Z}_n) = \mathbb{P}(|Z| \leqslant \epsilon_n) = 1-\mathbb{P}(|Z| > \epsilon_n) = 1$ which gives: $$\begin{align} \mathbb{P}(Z = 0) &= \mathbb{P} \bigg( \limsup_{n \rightarrow \infty} \mathscr{Z}_n \bigg) \\[6pt] &\geqslant \limsup_{n \rightarrow \infty} \mathbb{P} (\mathscr{Z}_n) \\[6pt] &= \limsup_{n \rightarrow \infty} 1 \\[6pt] &= 1, \\[6pt] \end{align}$$ which implies that $\mathbb{P}(Z=0)=1$. $\blacksquare$
Theorem: If $\mathbb{E}(|X-Y|)=0$ then $X^2=Y^2$ almost surely.
Proof: For any $\epsilon>0$ we can apply Markov's inequality to get:
$$\mathbb{P}(|X-Y|> \epsilon) \leqslant \frac{\mathbb{E}(|X-Y|)}{\epsilon} = 0,$$
which implies that $\mathbb{P}(|X-Y|=0)=1$ (see lemma above). We then have:
$$\begin{align} \mathbb{P}(X^2=Y^2) &= \mathbb{P}(|X|=|Y|) \\[6pt] &= \mathbb{P}(|X|-|Y|=0) \\[6pt] &\geqslant \mathbb{P}(|X-Y|=0) \\[6pt] &= 1, \\[6pt] \end{align}$$
which implies that $\mathbb{P}(X^2=Y^2)=1$ (i.e., $X^2=Y^2$ almost surely). $\blacksquare$
This question concerns a relationship between expectation and probability. Here is an elementary demonstration of the correct conclusion, which is that $X^2=Y^2$ almost surely: that is, the set on which $X^2\ne Y^2$ has zero probability.
Let $Z=|X-Y|$ and write $F_Z(z) = \Pr(Z\le z)$ for its distribution function. Since $F(z)=0$ when $z\lt 0,$ note that
$$0 = E[|X-Y|] = E[Z] = \int_{-\infty}^0 F_Z(z)\,\mathrm dz + \int_0^\infty (1-F_Z(z))\mathrm dz = \int_0^\infty (1-F_Z(z))\mathrm dz.$$
Because $1-F_Z(z) = \Pr(Z\gt z) \ge 0$ (the axiom of total probability), the right hand side implies $1-F_Z(z)=0$ almost surely on the domain $z\in [0,\infty);$ that is, because $F$ is non-decreasing, $F(z)=1$ for all $z\gt 0.$
Finally, because $F_Z$ is càdlàg,
$$\begin{aligned} \Pr(X^2=Y^2) &\ge \Pr(X=Y) \\ &= \Pr(|X-Y|=0) \\ &= \Pr(|X-Y| \le 0) \\& = F_Z(0) \\ &= \lim_{z\to 0^+} F_Z(z)\\ &= \lim_{z\to 0^+} 1 = 1, \end{aligned}$$
qed.
Property of Probability: If $\{A_n\}$ is a monotonic sequence increasing to $A$, then $P(A) = \lim_{n\to\infty} A_n$.
One has the algebraic identity $$|X^2-Y^2| = |X+Y||X-Y|\le (|X|+|Y|)|X-Y|$$
Since $\mathbb E[|X-Y|] =0 \implies |X-Y|=0 \text{ a.s}$. If furthermore $|X|,|Y|<\infty \text{ a.s.}$ it thus follows $|X^2-Y^2|=0 \text{ a.s.}$ by this inequality which is equivalent to $X^2 = Y^2\text{ a.s}$.
Note that $|X|,|Y|<\infty \text{ a.s.}$ must be implicitly assumed in the question. Indeed if we assumed $|X|=\infty$ on a set $A$ with positive measure then the requirement $\mathbb E[|X-Y|] = 0$ implies that on a positive measure subset $B\subseteq A$ we require $|\infty - Y| = 0$ which is ill defined. So the assumption $|X|,|Y|<\infty \text{ a.s.}$ must have been implicitly assumed in the question.
If you are allowed to use the standard fact that a measurable function vanishes almost everywhere iff the Lebesgue integral of its absolute value vanishes, then you can write
$0 = \mathbb E(|X-Y|) = \int |X-Y| \,\mathrm d \mathbb P$ to directly conclude
$1 = \mathbb P(X-Y=0) = \mathbb P(X=Y)$.
Since $\{X = Y\} \subseteq \{X^2 = Y^2\}$, monotonicity of $\mathbb P$ gives $1 = \mathbb P(X=Y) \leq \mathbb P(X^2 = Y^2)$ and hence $X^2 = Y^2$ with probability $1$.
