Convergence of random variables under different probability measures I have a succession of random variables $X_n$ on $\Omega=[0,1]$ with $X_n=(1-\omega)^n$.
I have to prove the convergence almost sure and/or in law in these case:


*

*$\mathbb P=\delta_{0}$

*$\mathbb P=\frac12 \delta_{0}+\frac12 \delta_{1}$

*$\mathbb P=$ Lebesgue measure on $[0,1]$


I dont' know how do.
 A: For 1), recall that $\mathbb P=\delta_0$ just means that $\mathbb P(\{0\})=1$. So $X_n(0) = (1-0)^n = 1$, and $\mathbb P(X_n=1)=1$. Hence trivially $X_n\stackrel{a.s.}{\longrightarrow}1$ and $X_n\stackrel{d}{\longrightarrow}1$
For 2), recall that $\mathbb P=\frac12\delta_0 + \frac12\delta_1$ means $\mathbb P(\{0\})=\mathbb P(\{1\})=\frac12$. So $X_n(0) = 1$ and $X_n(1)=0$, and $\mathbb P(X_n=0)=\mathbb P(X_n=1)=\frac12$. Since the $X_n$ all have the same distribution, $X_n\stackrel{a.s.}{\longrightarrow}X_1$ and $X_n\stackrel{d}{\longrightarrow}X_1$.
For 3), recall that Lebesgue measure on $[0,1]$ is just a uniform distribution over $[0,1]$, that is, $\mathbb P([a,b])=b-a$ for $0\leqslant a\leqslant b\leqslant 1$. Observe that 
$$(1-\omega)^n\stackrel{n\to\infty}{\longrightarrow}\begin{cases} 0,& \omega\ne 0\\1,&\omega=0.\end{cases}$$
Since $\mathbb P(\{0\})=0$, it follows that
$$\mathbb P\left(\lim_{n\to\infty} X_n=0 \right)=1, $$
i.e. $X_n\stackrel{a.s.}{\longrightarrow}0$ (and therefore $X_n\stackrel{d}{\longrightarrow}0$).
