[proofplan] We prove convergence in probability directly from the distribution-function characterization of [convergence in distribution](/page/Convergence%20In%20Distribution). For a fixed $\varepsilon>0$, the two points $c-\varepsilon$ and $c+\varepsilon$ are continuity points of the degenerate distribution at $c$. Convergence in distribution therefore forces the distribution functions of $X_n$ to converge to $0$ at $c-\varepsilon$ and to $1$ at $c+\varepsilon$, and these two one-sided tail estimates bound the probability of the event $|X_n-c|>\varepsilon$. [/proofplan] [step:Introduce the distribution functions of $X_n$ and of the degenerate limit] For each $n\in\mathbb N$, define the distribution function $F_n:\mathbb R\to [0,1]$ by $F_n(x)=\mathbb P(X_n\le x)$. Define the distribution function of the constant [random variable](/page/Random%20Variable) with value $c$ by $F:\mathbb R\to [0,1]$ given by $F(x)=\mathbb P(c\le x)$. Thus $F(x)=0$ for $x<c$ and $F(x)=1$ for $x\ge c$. Hence the only discontinuity point of $F$ is $c$. [/step] [step:Use convergence in distribution at the two endpoints around $c$] Fix $\varepsilon>0$. Since $c-\varepsilon<c<c+\varepsilon$, both $c-\varepsilon$ and $c+\varepsilon$ are continuity points of $F$. By the definition of $X_n\xrightarrow{d}c$ in terms of distribution functions, \begin{align*} \lim_{n\to\infty}F_n(c-\varepsilon)=F(c-\varepsilon)=0. \end{align*} Also, \begin{align*} \lim_{n\to\infty}F_n(c+\varepsilon)=F(c+\varepsilon)=1. \end{align*} [guided] We fix an arbitrary $\varepsilon>0$ because convergence in probability requires a separate estimate for every positive tolerance. The distribution function of the constant random variable with value $c$ is $F:\mathbb R\to [0,1]$ given by $F(x)=\mathbb P(c\le x)$. Since the event $\{c\le x\}$ is empty when $x<c$ and is all of $\Omega$ when $x\ge c$, we have $F(x)=0$ for $x<c$ and $F(x)=1$ for $x\ge c$. The jump of $F$ occurs only at $c$. Therefore $c-\varepsilon$ and $c+\varepsilon$ are continuity points of $F$. The definition of convergence in distribution says that at every continuity point $x$ of the limiting distribution function, \begin{align*} \lim_{n\to\infty}F_n(x)=F(x). \end{align*} Applying this first at $x=c-\varepsilon$ gives \begin{align*} \lim_{n\to\infty}F_n(c-\varepsilon)=F(c-\varepsilon)=0. \end{align*} Applying it again at $x=c+\varepsilon$ gives \begin{align*} \lim_{n\to\infty}F_n(c+\varepsilon)=F(c+\varepsilon)=1. \end{align*} These are exactly the two tail estimates needed to trap the probability that $X_n$ lies outside the interval $[c-\varepsilon,c+\varepsilon]$. [/guided] [/step] [step:Bound the probability of leaving the $\varepsilon$-neighbourhood of $c$] For each $n\in\mathbb N$, define the event \begin{align*} A_n:=\{\omega\in\Omega: |X_n(\omega)-c|>\varepsilon\}. \end{align*} If $\omega\in A_n$, then either $X_n(\omega)<c-\varepsilon$ or $X_n(\omega)>c+\varepsilon$. Hence \begin{align*} A_n\subset \{\omega\in\Omega:X_n(\omega)<c-\varepsilon\}\cup \{\omega\in\Omega:X_n(\omega)>c+\varepsilon\}. \end{align*} By [monotonicity and subadditivity](/theorems/1081) of $\mathbb P$, \begin{align*} \mathbb P(A_n)\le \mathbb P(X_n<c-\varepsilon)+\mathbb P(X_n>c+\varepsilon). \end{align*} Since $\{X_n<c-\varepsilon\}\subset\{X_n\le c-\varepsilon\}$, and since $\mathbb P(X_n>c+\varepsilon)=1-\mathbb P(X_n\le c+\varepsilon)$, we obtain \begin{align*} \mathbb P(|X_n-c|>\varepsilon)\le F_n(c-\varepsilon)+1-F_n(c+\varepsilon). \end{align*} [/step] [step:Let $n$ tend to infinity and conclude convergence in probability] Using the limits established above, \begin{align*} \lim_{n\to\infty}\bigl(F_n(c-\varepsilon)+1-F_n(c+\varepsilon)\bigr)=0. \end{align*} Because $0\le \mathbb P(|X_n-c|>\varepsilon)$ for every $n\in\mathbb N$, the preceding bound implies \begin{align*} \lim_{n\to\infty}\mathbb P(|X_n-c|>\varepsilon)=0. \end{align*} The number $\varepsilon>0$ was arbitrary, so $X_n\xrightarrow{\mathbb P}c$. [/step]