I'm trying to solve the following exercise:
Suppose that $X_n \to X$ in distribution. Show that $(X_n)_{n \geq 1}$ is a tight family.
First, just to recall the definition for those possibly unfamiliar, such a family of random variables (assumed to be defined on the same probability space, and real-valued) is tight iff for every $\epsilon > 0$, there is $K$ such that $\sup_n \mathbf{P}(|X_n| \geq K) \leq \epsilon$.
Here's the argument I had in mind. First let $1 > \epsilon > 0$ be given and notice that we can find a $K_0$ such that $F_X(K_0) \geq 1 – \epsilon/2$, since $F_X(t) \to 1$ as $t \to +\infty$. Without loss, we can also assume that $K_0$ is a continuity point of $F_X$ (simply because $F_X$ is nondecreasing, and there are countably many discontinuities of $F_X$, and so replace $K_0$ by some element of $[K_0, K_0 + 1]$, if necessary).
Notice that
$$
\mathbf{P}(|X_n| \geq K_0) = 1 – F_{X_n}(K_0) + F_X(K_0) – F_X(K_0) \leq (1 – F_X(K_0)) + |F_{X_n}(K_0) – F_X(K_0)|.
$$
By choice of $K_0$, $1 – F_X(K_0) \leq \epsilon/2$.
Hence since $X_n \to X$ in distribution and $K_0$ is a continuity point of $F_X$, there is $N$ large such that $\sup_{n \geq N} \mathbf{P}(|X_n| \geq K_0) \leq \epsilon$. For $n < N$, put $K_n$ such that $\mathbf{P}(|X_n| \geq K_n) \leq \epsilon$ (as before, such $K_n$ necessarily exists). Now taking $K= \max\{K_0, K_1, \dots, K_N\}$, we have $\sup_{n} \mathbf{P}(|X_n| \geq K) \leq \epsilon$, as needed.
Best Answer
The proof is correct. The only (small) typo is that we have to choose $K_n$ such that $\mathbf{P}(|X_n| \geq K_n) \leq \epsilon/2$ (not $\varepsilon$) in order to be sure that $\mathbf{P}(|X_n| \geq K_n) \leq \epsilon$. Otherwise, we would have $\mathbf{P}(|X_n| \geq K_n) \leq 3\epsilon/2$.