Proof of Relations Between the Modes of Convergence

Claim 1 (almost sure implies in probability). Fix $\varepsilon>0$ and let $C_k=\bigcup_{m\ge k}\{|X_m-X|\ge\varepsilon\}$, events by Sigma-Algebra and Measurable Space since each $|X_m-X|$ is a random variable (Step 0(a) of the proof of Linearity and Monotonicity of the Lebesgue Integral). The $C_k$ decrease, and every $\omega\in\bigcap_k C_k$ satisfies $|X_m(\omega)-X(\omega)|\ge\varepsilon$ for infinitely many $m$, so $X_m(\omega)\not\to X(\omega)$; hence $\bigcap_k C_k$ is contained in the complement of the convergence event of Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution, which has probability $0$ by hypothesis. By continuity from above of probability measures (Preliminaries of the proof of Borel-Cantelli Lemmas), $P(C_k)\to 0$; and monotonicity gives $P(|X_k-X|\ge\varepsilon)\le P(C_k)\to 0$.

Claim 2 (in probability implies in distribution). Fix $t\in\mathbb{R}$ at which $F_X$ is continuous, and let $\varepsilon>0$. If $X_m\le t$ then either $X\le t+\varepsilon$ or $|X_m-X|>\varepsilon$; by finite subadditivity of $P$,

Symmetrically, if $X\le t-\varepsilon$ then either $X_m\le t$ or $|X_m-X|>\varepsilon$, so

Letting $m\to\infty$ with $\varepsilon$ fixed, every subsequential behavior of $F_{X_m}(t)$ is confined to the interval $[F_X(t-\varepsilon),F_X(t+\varepsilon)]$ in the sense that

with $\liminf$ and $\limsup$ of bounded real sequences as in the proof of Dominated Convergence Theorem. Letting $\varepsilon\to 0$ and using continuity of $F_X$ at $t$, both bounds tend to $F_X(t)$, so $F_{X_m}(t)\to F_X(t)$, as required by Almost Sure Convergence, Convergence in Probability, and Convergence in Distribution.

Claim 3 (convergence in distribution to a constant). Let $F_c$ denote the cumulative distribution function of the constant $c$, so $F_c(t)=0$ for $t<c$ and $F_c(t)=1$ for $t\ge c$, and $F_c$ is continuous at every $t\ne c$. Fix $\varepsilon>0$. Then

using $\{X_m\ge c+\varepsilon\}\subseteq\{X_m>c+\varepsilon/2\}$ and finite subadditivity. The points $c-\varepsilon$ and $c+\varepsilon/2$ are continuity points of $F_c$, so by hypothesis $F_{X_m}(c-\varepsilon)\to F_c(c-\varepsilon)=0$ and $F_{X_m}(c+\varepsilon/2)\to F_c(c+\varepsilon/2)=1$; hence the right side tends to $0$, proving convergence in probability to $c$. $\blacksquare$