Proof.

Quadratic mean implies probability

By Markov’s inequality applied to \(\left (X_n-X\right )^{2}\), for any \(\varepsilon >0\), \[P\left (\left |X_n-X\right |\geq \varepsilon \right ) = P\left [\left (X_n-X\right )^{2}\geq \varepsilon ^{2}\right ] \leq \frac {E\left [\left (X_n-X\right )^{2}\right ]}{\varepsilon ^{2}} \longrightarrow 0 .\]

Probability implies distribution

This is part (3) of the limit theorems of the next chapter, proved there by squeezing \(F_{X_n}(x)\) between \(F_X(x\pm \varepsilon )\) and letting \(\varepsilon \downarrow 0\) at a continuity point.

Almost sure implies probability

Fix \(\varepsilon >0\) and set \(A_n = \bigcup _{m\geq n}\left \{\left |X_m-X\right |\geq \varepsilon \right \}\). The sets \(A_n\) decrease, and on the event \(\{X_n\rightarrow X\}\) — which has probability one — every outcome lies outside \(A_n\) for all large \(n\). Hence \(P\left (\bigcap _n A_n\right )=0\), and by continuity of measure from above \(P(A_n)\rightarrow 0\). Since \(\left \{\left |X_n-X\right |\geq \varepsilon \right \}\subseteq A_n\), the probability of the former tends to zero as well.

Distribution to a constant implies probability

Suppose \(X_n\underset {D}{\longrightarrow }c\). The limiting distribution function is the step at \(c\), continuous everywhere except at \(c\) itself, so for any \(\varepsilon >0\) \[P\left (\left |X_n-c\right |\geq \varepsilon \right ) \leq P\left (X_n \leq c-\varepsilon \right ) + P\left (X_n > c+\tfrac {\varepsilon }{2}\right ) = F_{X_n}\left (c-\varepsilon \right ) + 1 - F_{X_n}\left (c+\tfrac {\varepsilon }{2}\right ),\] and both terms converge to \(0 + 1 - 1 = 0\), the points \(c\pm \) being continuity points of the limit. □