Proving if $T_n$ converges in distribution to constant $c$, then $T_n$ converges in probability to $c$

Question

Show that if $T_n \xrightarrow[]{\text{d}} c$, then $T_n \xrightarrow[]{\text{c}} c$.

I have tried proving this, and got stuck. Here were my steps:

$\lim_{n\to\infty} P(|T_n - C| < \epsilon)$

$= \lim_{n\to\infty} P(-\epsilon < T_n - C < \epsilon)$

$= \lim_{n\to\infty} P(c-\epsilon < T_n < c+ \epsilon)$

$= \lim_{n\to\infty} [F_{T_n}(c+\epsilon) - F_{T_n}(c-\epsilon)]$

I don't know how to proceed from this point onwards. What piece of information am I possibly missing?

You need to express you condition. What does $T_n \xrightarrow[]{\text{d}} c$ imply? — leonbloy, Aug 19 '21 at 23:32
Specifically, what does $T_n \overset{d}{\to} c$ say about the CDFs $F_{T_n}$? — angryavian, Aug 19 '21 at 23:55
Let me know if the proof given in the question post of Proof for convergence in distribution implying convergence in probability for constants answers your question. — Sarvesh Ravichandran Iyer, Aug 23 '21 at 16:54

score 1 · Accepted Answer · answered Aug 20 '21 at 03:25

The CDF of the constant $c$ is $F(x) = \begin{cases}1 & x \ge c \\ 0 & x < c \end{cases}$.

One definition of $T_n \overset{d}{\to} c $ is that $F_{T_n}(x) \to F(x)$ for $x$ at which $F$ is continuous. Use this to compute $\lim_{n \to \infty} F_{T_n}(c+\epsilon)$ and $\lim_{n \to \infty} F_{T_n}(c-\epsilon)$.

Proving if $T_n$ converges in distribution to constant $c$, then $T_n$ converges in probability to $c$

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