[proofplan] We prove the three implications separately. Almost sure convergence is converted into convergence in probability by applying dominated convergence to the indicators of the deviation events. $L^p$ convergence implies convergence in probability by Markov's inequality applied to $|X_n-X|^p$. Finally, convergence in probability implies convergence in distribution by comparing the distribution functions of $X_n$ and $X$ at continuity points of the distribution function of $X$. [/proofplan] [step:Declare the ambient probability space and the convergence notions] Let $\mathbb N:=\{1,2,3,\ldots\}$. Let $(\Omega,\mathcal F,\mathbb P)$ be the probability space, let $\mathcal B(\mathbb R)$ denote the Borel $\sigma$-algebra on $\mathbb R$, and let \begin{align*} X_n:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)),\qquad n\in\mathbb N,\\ X:(\Omega,\mathcal F)&\to(\mathbb R,\mathcal B(\mathbb R)) \end{align*} be the real-valued random variables in the theorem. For a real number $\varepsilon>0$, convergence in probability $X_n\xrightarrow{\mathbb P}X$ means \begin{align*} \lim_{n\to\infty}\mathbb P(\{\omega\in\Omega:|X_n(\omega)-X(\omega)|>\varepsilon\})=0. \end{align*} Almost sure convergence $X_n\xrightarrow{a.s.}X$ means that there exists an event $\Omega_0\in\mathcal F$ with $\mathbb P(\Omega_0)=1$ such that \begin{align*} \lim_{n\to\infty}X_n(\omega)=X(\omega) \end{align*} for every $\omega\in\Omega_0$. For a real number $p\ge1$, convergence $X_n\xrightarrow{L^p}X$ means that $|X_n-X|^p$ is integrable for each $n\in\mathbb N$ and \begin{align*} \lim_{n\to\infty}\mathbb E[|X_n-X|^p] = \lim_{n\to\infty}\int_\Omega |X_n(\omega)-X(\omega)|^p\,d\mathbb P(\omega) =0. \end{align*} For a real-valued random variable $Y:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$, define its distribution function \begin{align*} F_Y:\mathbb R&\to[0,1]\\ t&\mapsto \mathbb P(\{\omega\in\Omega:Y(\omega)\le t\}). \end{align*} Convergence in distribution $X_n\xrightarrow{d}X$ means that \begin{align*} \lim_{n\to\infty}F_{X_n}(t)=F_X(t) \end{align*} for every real number $t\in\mathbb R$ at which $F_X$ is continuous. [/step] [step:Convert almost sure convergence into convergence in probability] Assume $X_n\xrightarrow{a.s.}X$. Fix a real number $\varepsilon>0$. For each $n\in\mathbb N$, define the deviation event \begin{align*} A_n^\varepsilon:=\{\omega\in\Omega:|X_n(\omega)-X(\omega)|>\varepsilon\}\in\mathcal F \end{align*} and define the indicator random variable \begin{align*} \mathbb 1_{A_n^\varepsilon}:\Omega&\to\{0,1\}\\ \omega&\mapsto \begin{cases} 1,&\omega\in A_n^\varepsilon,\\ 0,&\omega\notin A_n^\varepsilon. \end{cases} \end{align*} Since $X_n\to X$ almost surely, $\mathbb 1_{A_n^\varepsilon}(\omega)\to0$ for $\mathbb P$-almost every $\omega\in\Omega$. Also $0\le \mathbb 1_{A_n^\varepsilon}\le1$ on $\Omega$, and the constant function $1:\Omega\to\mathbb R$ is integrable because $\mathbb P(\Omega)=1$. By the [Dominated Convergence Theorem](/theorems/???), \begin{align*} \lim_{n\to\infty}\mathbb P(A_n^\varepsilon) = \lim_{n\to\infty}\int_\Omega \mathbb 1_{A_n^\varepsilon}(\omega)\,d\mathbb P(\omega) = \int_\Omega 0\,d\mathbb P(\omega) = 0. \end{align*} Because this holds for every $\varepsilon>0$, $X_n\xrightarrow{\mathbb P}X$. [/step] [step:Use Markov's inequality to pass from $L^p$ convergence to convergence in probability] Assume $X_n\xrightarrow{L^p}X$ for some real number $p\ge1$. Fix a real number $\varepsilon>0$. For each $n\in\mathbb N$, define \begin{align*} Y_n:\Omega&\to[0,\infty)\\ \omega&\mapsto |X_n(\omega)-X(\omega)|^p. \end{align*} The random variable $Y_n$ is nonnegative and integrable by the definition of $L^p$ convergence. Applying [Markov's Inequality](/theorems/???) to $Y_n$ with threshold $\varepsilon^p>0$ gives \begin{align*} \mathbb P(\{\omega\in\Omega:|X_n(\omega)-X(\omega)|>\varepsilon\}) &= \mathbb P(\{\omega\in\Omega:Y_n(\omega)>\varepsilon^p\})\\ &\le \frac{\mathbb E[Y_n]}{\varepsilon^p}\\ &= \frac{1}{\varepsilon^p}\int_\Omega |X_n(\omega)-X(\omega)|^p\,d\mathbb P(\omega). \end{align*} Taking $n\to\infty$ and using $X_n\xrightarrow{L^p}X$ yields \begin{align*} \lim_{n\to\infty}\mathbb P(\{\omega\in\Omega:|X_n(\omega)-X(\omega)|>\varepsilon\})=0. \end{align*} Because this holds for every $\varepsilon>0$, $X_n\xrightarrow{\mathbb P}X$. [/step] [step:Compare distribution functions at continuity points] Assume $X_n\xrightarrow{\mathbb P}X$. Let $t\in\mathbb R$ be a point at which the distribution function $F_X:\mathbb R\to[0,1]$ is continuous. Fix a real number $\varepsilon>0$. For each $n\in\mathbb N$, the event $\{X_n\le t\}$ is contained in the union of the events $\{X\le t+\varepsilon\}$ and $\{|X_n-X|>\varepsilon\}$, because if $X_n(\omega)\le t$ and $|X_n(\omega)-X(\omega)|\le\varepsilon$, then $X(\omega)\le t+\varepsilon$. Hence \begin{align*} F_{X_n}(t) = \mathbb P(\{X_n\le t\}) \le F_X(t+\varepsilon)+\mathbb P(\{|X_n-X|>\varepsilon\}). \end{align*} Taking the limit superior and using convergence in probability gives \begin{align*} \limsup_{n\to\infty}F_{X_n}(t)\le F_X(t+\varepsilon). \end{align*} Similarly, the event $\{X\le t-\varepsilon\}$ is contained in the union of the events $\{X_n\le t\}$ and $\{|X_n-X|>\varepsilon\}$, because if $X(\omega)\le t-\varepsilon$ and $|X_n(\omega)-X(\omega)|\le\varepsilon$, then $X_n(\omega)\le t$. Therefore \begin{align*} F_X(t-\varepsilon) \le F_{X_n}(t)+\mathbb P(\{|X_n-X|>\varepsilon\}). \end{align*} Taking the limit inferior and using convergence in probability gives \begin{align*} F_X(t-\varepsilon)\le \liminf_{n\to\infty}F_{X_n}(t). \end{align*} [guided] We prove convergence of distribution functions at a fixed continuity point $t$ of $F_X$. The goal is to trap $F_{X_n}(t)$ between values of $F_X$ slightly to the left and slightly to the right of $t$, with an error term controlled by convergence in probability. Fix a real number $\varepsilon>0$. For each $n\in\mathbb N$, consider the event \begin{align*} \{\omega\in\Omega:X_n(\omega)\le t\}. \end{align*} If, in addition, $|X_n(\omega)-X(\omega)|\le\varepsilon$, then \begin{align*} X(\omega)\le X_n(\omega)+\varepsilon\le t+\varepsilon. \end{align*} Thus every $\omega\in\Omega$ with $X_n(\omega)\le t$ must satisfy at least one of the two alternatives $X(\omega)\le t+\varepsilon$ or $|X_n(\omega)-X(\omega)|>\varepsilon$. This proves the event inclusion \begin{align*} \{\omega\in\Omega:X_n(\omega)\le t\} \subset \{\omega\in\Omega:X(\omega)\le t+\varepsilon\} \cup \{\omega\in\Omega:|X_n(\omega)-X(\omega)|>\varepsilon\}. \end{align*} Taking probabilities and using subadditivity of the probability measure $\mathbb P$ gives \begin{align*} F_{X_n}(t) = \mathbb P(\{\omega\in\Omega:X_n(\omega)\le t\}) \le F_X(t+\varepsilon)+\mathbb P(\{\omega\in\Omega:|X_n(\omega)-X(\omega)|>\varepsilon\}). \end{align*} Since $X_n\xrightarrow{\mathbb P}X$, the second term on the right tends to $0$ as $n\to\infty$. Therefore \begin{align*} \limsup_{n\to\infty}F_{X_n}(t)\le F_X(t+\varepsilon). \end{align*} For the lower bound, we compare from the left. If $X(\omega)\le t-\varepsilon$ and $|X_n(\omega)-X(\omega)|\le\varepsilon$, then \begin{align*} X_n(\omega)\le X(\omega)+\varepsilon\le t. \end{align*} Thus every $\omega\in\Omega$ with $X(\omega)\le t-\varepsilon$ must either satisfy $X_n(\omega)\le t$ or satisfy $|X_n(\omega)-X(\omega)|>\varepsilon$. Hence \begin{align*} \{\omega\in\Omega:X(\omega)\le t-\varepsilon\} \subset \{\omega\in\Omega:X_n(\omega)\le t\} \cup \{\omega\in\Omega:|X_n(\omega)-X(\omega)|>\varepsilon\}. \end{align*} Taking probabilities and using subadditivity gives \begin{align*} F_X(t-\varepsilon) \le F_{X_n}(t)+\mathbb P(\{\omega\in\Omega:|X_n(\omega)-X(\omega)|>\varepsilon\}). \end{align*} Again convergence in probability makes the error term tend to $0$, so \begin{align*} F_X(t-\varepsilon)\le \liminf_{n\to\infty}F_{X_n}(t). \end{align*} These two inequalities are the essential comparison: probability convergence controls the error event, while continuity of $F_X$ will let $\varepsilon$ shrink to $0$ in the next step. [/guided] [/step] [step:Let the comparison width shrink to zero and conclude convergence in distribution] From the previous step, for every real number $\varepsilon>0$, \begin{align*} F_X(t-\varepsilon) \le \liminf_{n\to\infty}F_{X_n}(t) \le \limsup_{n\to\infty}F_{X_n}(t) \le F_X(t+\varepsilon). \end{align*} Because $F_X$ is continuous at $t$, the one-sided limits satisfy \begin{align*} \lim_{\varepsilon\downarrow0}F_X(t-\varepsilon)=F_X(t), \qquad \lim_{\varepsilon\downarrow0}F_X(t+\varepsilon)=F_X(t). \end{align*} Letting $\varepsilon\downarrow0$ in the preceding inequalities gives \begin{align*} F_X(t) \le \liminf_{n\to\infty}F_{X_n}(t) \le \limsup_{n\to\infty}F_{X_n}(t) \le F_X(t). \end{align*} Therefore \begin{align*} \lim_{n\to\infty}F_{X_n}(t)=F_X(t). \end{align*} Since $t$ was an arbitrary continuity point of $F_X$, this is exactly $X_n\xrightarrow{d}X$. Combining the three separately proved implications completes the proof. [/step]