Combining converge in distribution with convergence in probability Let $\left\{ T_{n}:n\geq1\right\}$  be a sequence of random variables such that $T_{n}\overset{d}{\to}T$ as $n\to\infty$, that is $P\left\{ T_{n}\leq t\right\} \to P\left\{ T\leq t\right\}$  for all $t$ at which $P\left\{ T\leq t\right\}$  is continuous. Let $\left\{ C_{n}:n\geq1\right\}$  be a sequence of random variables such that $C_{n}\overset{P}{\to}c$ as $n\to\infty$, that is for $\varepsilon>0$
$P\left\{ \left|C_{n}-c\right|>\varepsilon\right\} \to0$ as $n\to\infty$.
Suppose that $P\left\{ T\leq t\right\}$  is continuous at $c$. I want to show that $P\left\{ T_{n}\leq C_{n}\right\} \to P\left\{ T\leq c\right\}$  as $n\to\infty$. Also, I want to further show by example that the result may not be true without the continuity assumption.
I honestly I have no idea where to begin to prove this.
 A: I hope I have done the calculations correctly. We derive two simple inclusions:
$$
\{ T_n \le c - \epsilon \} \subset \{ T_n \le C_n \} \cup \{ C_n \le c- \epsilon\}
$$
and 
$$
\{ T_n \le C_n \} \subset \{ c+\epsilon \le C_n \} \cup \{T_n \le c+ \epsilon\}
$$
Now we can immediately find the bound:
$$
P( T_n \le c - \epsilon_1) - P(C_n \le c- \epsilon_1) \le P(T_n \le C_n) \le P( T_n \le c + \epsilon_2) + P(C_n \ge c+ \epsilon_2)
$$
Now since there can be only a countable number of discontinuities in the CDF we can chose sequences $\epsilon_{1,k}$ and $\epsilon_{2,k}$ such that they converge to zero and $c-\epsilon_{1,k}$ and $c+\epsilon_{2,k}$ are continuity points for the CDF of $T$. so we find that:
$$
P( T \le c - \epsilon_{1,k})  \le \liminf_n P(T_n \le C_n) \le \limsup_n P(T_n \le C_n)\le P( T \le c ) 
$$
where on the right-hand side we used the right continuity of the CDF. Now all that is left to do is use the continuity of the CDF in $c$ to pass to the limit on the left-hand side. This gives us the result.
As for the counterexample, take $T_n = 0$ deterministic and $C_n = -\frac{1}{n}.$ Then 
$$ 0 = \lim_n P(T_n \le C_n) < P(T \le 0) = 1.$$
