Showing $X_n \rightarrow X$ and $X_n \rightarrow Y$ implies $X\overset{\text{a.s.}}{=}Y$ for four types of convergence

I need to show that if $$X_n \rightarrow X$$ and $$X_n \rightarrow Y$$, then $$X\overset{\text{a.s.}}{=}Y$$ for convergence in probability, convergence almost surely, as well as for convergence in mean and quadratic mean ($$\mathcal L^1$$ and $$\mathcal L^2$$ convergence).

Convergence in Probability:

For any $$\epsilon>0$$ and for any $$n\in\mathbb N$$ we have

\begin{align} \mathbb P(|X-Y|\geq\epsilon) &\leq\mathbb P(|X-X_n|+|X_n-Y|\geq\epsilon)\\\\ &\leq\mathbb P\left((|X-X_n|\geq\epsilon/2)\cup(|X_n-Y|\geq\epsilon/2)\right)\\\\ &\leq\mathbb P(|X-X_n|\geq\epsilon/2)+\mathbb P(|X_n-Y|\geq\epsilon/2) \end{align}

so that

$$\mathbb P(|X-Y|\geq\epsilon)\leq\lim_{n\rightarrow\infty}\mathbb P(|X-X_n|\geq\epsilon/2)+\mathbb P(|X_n-Y|\geq\epsilon/2)=0$$

Since $$\{|X-Y|>0\}=\underbrace{\bigcup_{n=1}^\infty \underbrace{\left\{|X-Y|>\frac{1}{n}\right\}}_{=\emptyset}}_{=\emptyset}=\emptyset$$

we have that $$\mathbb P\{|X-Y|>0\}=0$$ and so $$\mathbb P(X\ne Y)=0$$. Hence $$\mathbb P(X= Y)=1$$ which means that $$X\overset{\text{a.s.}}{=}Y$$.

Convergence Almost Surely:

Since almost sure convergence implies convergence in probability, the result follows immediately from the last part. However, I'd like to show this without making use of that result. Since $$X_n$$ converges almost surely to both $$X$$ and $$Y$$ then $$\mathbb P(\lim_{n\rightarrow\infty}X_n=X)=1$$ and $$\mathbb P(\lim_{n\rightarrow\infty}X_n=Y)=1$$. From here it seems obvious to me that $$X\overset{\text{a.s.}}{=}Y$$ but I'm not sure how to show this formally.

Convergence in Mean:

\begin{align} \mathbb E(|X-Y|) &\leq\mathbb E\left(|X-X_n|+|X_n-Y|\right)\\\\ &=\mathbb E\left(|X-X_n|)+\mathbb E(|X_n-Y|\right) \end{align}

so

$$\mathbb E(|X-Y|)\leq\lim_{n\rightarrow\infty}\mathbb E(|X-X_n|)+\mathbb E(|X_n-Y|)=0$$

so $$X\overset{\text{a.s.}}{=}Y$$

I tried continuing with the same logic but it's not the case that

$$\mathbb E(|X-Y|^2)\leq\mathbb E\left(|X-X_n|^2+|X_n-Y|^2\right)$$

so I'm not sure how to proceed.

Is my reasoning correct for the first and third? How can I proceed with the other two?

• Yes, first and third seems correct. For the second, note that $X_n \to X, X_n \to Y$ almost surely, means you have a set $A$ of full measure, such that for $\omega \in A$ you have $X_n(\omega) \to X(\omega)$ and $X_n(\omega) \to Y(\omega)$ as a sequence of numbers, so easily $|X(\omega)-Y(\omega)| \le |X(\omega)-X_n(\omega)| + |Y(\omega) - X_n(\omega)| \to 0$ which means that you have $X=Y$ on $A$, hence almosts surely. For the fourth: Minkowsky Inequality and goes the same as third one Oct 4 '20 at 19:58

\begin{align*} \left(\mathbb E\left(|X-Y|^2\right)\right)^{1/2} &=\left(\mathbb E\left(|(X-X_n)+(X_n-Y)|^2\right)\right)^{1/2}\\\\ &\leq \left(\mathbb E\left(|X-X_n|^2\right)\right)^{1/2}+\left(\mathbb E\left(|X_n-Y|^2\right)\right)^{1/2} \end{align*}
\begin{align*} \left(\mathbb E\left(|X-Y|^2\right)\right)^{1/2} &\leq\lim_{n\rightarrow\infty} \left(\mathbb E\left(|X-X_n|^2\right)\right)^{1/2}+\left(\mathbb E\left(|X_n-Y|^2\right)\right)^{1/2}\\\\ &=0 \end{align*}
so $$X\overset{\text{a.s.}}{=}Y$$
It follows from convergence a.s. and from convergence in $$L_p$$ that there is convergence in probability. So it was enough to condsider convergence in probability.