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Let $X, X_1,X_2,\ldots$ be random variables that are defined on the same probability space $(\Omega, \mathcal{F}, \mathbb{P})$. Suppose that $X_n \xrightarrow{\mathbb{P}} X$ and that there exists a radom variables $Y$ with $\mathbb{E}Y < \infty$ such that $|X_n|\leq Y$.

I want to show that $\mathbb{P}(|X|\leq Y)=1$ and that $X_n \xrightarrow{\mathcal{L}^1} X$.

The latter follows easily since $Y \in \mathcal{L}^1$, so for all $n \in \mathbb{N}$ \begin{align} &\lim_{K \to \infty} \sup \{\mathbb{E} |X_n| \mathbb{1}_{\{|X_n| > K \}} \} \\ &\leq \lim_{K \to \infty} \mathbb{E} Y \mathbb{1}_{\{Y > K \}}=0. \end{align} It follows that $(X_n)_{n \in \mathbb{N}}$ is uniformly integrable as well. Therefore, $X_n \xrightarrow{\mathcal{L}^1} X$.

However, I don't know how to prove the first statement. Just by starting from the definition of convergence in probability right away? \begin{align} \forall \epsilon: \lim_{n \to \infty} \mathbb{P}(|X_n - X|>\epsilon)=0 \end{align} Using the triangle-inequality does not seem to be that fruitful. Any help is appreciated.

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    @ iJup: you could also say that convergence in probability implies that there exists a subsequence $X_{n_k}$ converging a.s. to $X$. Therefore $|X_{n_k}| \to |X|$ a.s. As $|X_{n_k}| \leq Y$, then $|X| \leq Y$. – 2017-01-02

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Let $M\in\mathbb{N}$ be arbitrary.

Since $\lim\limits_{n\to \infty}\mathbb{P}(|X-X_n|>1/M)=0$ and $|X_n|\leq Y$ we have that $\mathbb{P}(|X|>Y+1/M) = 0.$ More explicitly, let $\delta>0$ be arbitrary, then there exists a $N$ large enough such that $\mathbb{P}(|X-X_N|>1/M)<\delta$. Then \begin{align*} \mathbb{P}(|X|>Y+1/M)\leq& \mathbb{P}(|X|>|X_N|+1/M) \\ \leq & \mathbb{P}(|X-X_N|>1/M) \\ <& \delta. \end{align*}As $\delta$ arbitrary we have $\mathbb{P}(|X|>Y+1/M) = 0.$ However, notice that the sets $\{ |X|>Y+1/M \}_{M}$ are increasing. Hence we have \begin{align*} \mathbb{P}(|X|> Y) &= \mathbb{P}(\bigcup_{M=1}^\infty\{ |X|>Y+1/M \}) \\ &= \lim\limits_{M\to \infty} \mathbb{P}(|X|>Y+1/M) =0. \end{align*}

Hence $\mathbb{P}(|X|\leq Y)=1.$