Prove that a sequence defined by $Y_i = \sqrt{X_i}$ converges in probability to $\sqrt{a}$ Suppose a sequence of random variables $X_1 ... X_n$ converges in probability to a constant $a$. Furthemore, suppose $\mathbb{P}[X_i > 0] = 1 \ \forall \ i$, i.e., $X_i$ are all positive random variables almost surely. Prove that a sequence defined by $Y_i = \sqrt{X_i}$ converges in probability to $\sqrt{a}$. 
My attempt: since $\frac{1}{\sqrt n} \sum_{k=1}^n X_i = \sqrt n \left(\frac 1n \sum_{k=1}^n X_i\right)$, $\left(\frac{1}{\sqrt n} \sum_{k=1}^n X_i\right)_n$ converges in distribution to \mathcal $N(0,\sigma^2)$, so the question essentially asks to prove that convergence cannot be improved to convergence in probability.
Supposing for the sake of contradiction that $\left(\frac{1}{\sqrt n} \sum_{k=1}^n X_i\right)_n$ converges to some $Y$ in probability, there exists a subsequence that converges to $Y$ almost surely. But what to do next I do not know.
 A: WLOG, assume all members of the sequence are positive. 
Then use that if $X_n\to X$ in probability and $g:\mathbb{R}\to \mathbb{R}$ is continuous, then $g(X_n)\to g(X)$ in probability.
To prove this last fact, use the relation of convergence in probability with subsequences that converge almost surely.
Also, in your attempt you used CLT, which cannot be applied (you don't know anything about the independence of the sequence nor the distributions nor the moments).
A: Here is a proof based only on the definition of convergence in probability.
I will assume that $a>0$ and leave the (trivial)  case $a=0$ to you.
Let $0 <\epsilon <a$. $$P(|\sqrt {X_i} -\sqrt a| >\epsilon)$$ $$ =P(\frac  {|X_i-a|} {|\sqrt {X_i} +\sqrt a|} >\epsilon)$$ $$=P(|X_i-a| > \epsilon (|\sqrt {X_i} +\sqrt a|).$$ 
Now split this into the parts where $|X_i -a| >\epsilon$ and $|X_i -a| \leq \epsilon$. For the second part note that $$P(|\sqrt {X_i} -\sqrt a| >\epsilon, |X_i-a| <\epsilon)$$ $$ \leq P(|X_i-a| > \epsilon (|\sqrt {a-\epsilon} +\sqrt a|).$$ Now it is easy to finish the proof. 
