Let $Y ∼ U[0, 1]$, and let $X_n = Y^n$. Prove that $X_n\to 0$ in probability. Let $Y ∼ U[0, 1]$, and let $X_n = Y^n$. Prove that $X_n\to 0$ in probability. 
I guess I need to use $\lim_{n\to\infty} P(|X_n − Y| ≥ \epsilon) = 0.
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 A: Since $0<Y<1$ almost surely we have $X_n=Y^{n} \to 0$ almost surely. Convergence with probability $1$ implies convergence in probability. Aliter: $P\{X_n > \epsilon\}=P\{Y >\epsilon^{1/n}\}=1-\epsilon^{1/n} \to 1-1= 0$ as $n \to \infty$.
A: You want to show that $X_n$ converges to $0$ and not to $Y$, so you need to show that $$P(|X_n-0|\ge\epsilon)\to 0$$ 
and not $P(|X_n-Y|\ge \epsilon)\to0$. You have that 
$$P(|X_n-0|\ge \epsilon)=P(X_n\ge \epsilon)=P(Y^n\ge \epsilon)=P(Y\ge \epsilon^{1/n})=\int_{\epsilon^{1/n}}^1dt=1-\epsilon^{1/n}$$
A: You need to prove that for any $\varepsilon>0$ we have
\begin{align*}
\lim_{n\to\infty} \mathbb P[Y^n>\varepsilon]=0
\end{align*}
Observe that
\begin{align*}
\lim_{n\to\infty} \mathbb P[Y^n>\varepsilon]&=\lim_{n\to\infty} \mathbb P[Y>\varepsilon^{\frac{1}{n}}]\\
&=1-\lim_{n\to\infty} F_{Y} (\varepsilon^{\frac{1}{n}})\\
&=1-F_Y(1)\\
&=0
\end{align*}
Which proves convergence in distribution. In this case we can also prove almost sure convergence because for any $\omega\in\Omega$, $\lim_{n\to\infty}Y(\omega)^n=0$ unless $Y(\omega)=1$ so
\begin{align*}
\mathbb P[\lim_{n\to\infty} Y^n = 0] &= 1-\mathbb{P}[Y=1]\\
&=1
\end{align*}
Almost sure convergence gives you convergence in probability.
