I would like to prove the following statement:
Let $s_n(X_n - c)$ converge in distribution to $F$, where $F$ is the cumulative distribution of a tight random variable T, $s_n$ is a sequence of real numbers such that $s_n \to \infty$, and $c$ is a constant. Then $X_n$ converges in probability to $c$.
What I tried is starting with the definition of convergence in distribution. Let $\epsilon > 0$. We know that
$\lim \limits_{n \to \infty} P(s_n(X_n - c) \leq x) = P(T \leq x)$
and that since $T$ is tight, we know that there exists a real number $M > 0$ such that $P(|T| \leq x) \geq 1 - \epsilon$ for all $\epsilon > 0$. I tried writing
$P(s_n(X_n - c)) = P(X_n - c \leq \frac{x}{s_n}$),
of which we know that $\frac{x}{s_n}$ goes to zero and becomes smaller than $\epsilon$ as $n \to \infty$. We would like to show that
$\lim\limits_{n \to \infty} P(X_n - c < \epsilon) = 1$ for all $\epsilon > 0$, but unfortunately I do not see how the fact that $T$ is tight helps me in this case, or how to continue from my last step.
Any help would greatly be appreciated.
Suppose not. Then there are $\epsilon>0$ and a subsequence $n_k$ s.t.
$$ \mathsf{P}(|X_{n_k}-c|>\epsilon)\to \eta>0. $$
But then for any $M>0$ and $k$ large enough,
$$ \mathsf{P}(s_{n_k}|X_{n_k}-c|>M)\ge \mathsf{P}(s_{n_k}|X_{n_k}-c|>s_{n_k}\epsilon)\to\eta $$
and the LHS of the last expression converges to $\mathsf{P}(|T|>M)$ (assuming that $M$ and $-M$ are continuity points of $F$).