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The Exponential martingale for Brownian motion is a true martingale: independent increments and the normal distribution identity $\mathbb{E}{e}^{aN}=e^{a^{2}v/2}$ for $N\sim N(0,v)$ give $\mathbb{E}[Z_t\mid {\mathcal{F}}_s]=Z_s$. Its second moment isThe uniform integrability from bounded second moments criterion now applies: for $K>0$,Thus $\{Z_t:0\le t\le T\}$ has uniform integrability. In fact, all values stopped at stopping times bounded by $T$ have uniform integrability, since $Z_{\tau }=\mathbb{E}[Z_T\mid {\mathcal{F}}_{\tau }]$ and uniform integrability of conditional expectations applies. This is a finite-horizon assertion, not an assertion of uniform integrability over all $t\ge 0$.

From convergence in probability, choose a subsequence $X_{n_k}\to X$ almost surely. The Fatou lemma then givesso $X\in L^2$.

The bounded second moments make $(X_n)$ uniformly integrable, and $X$ is integrable. Convergence in probability together with uniform integrability therefore givesConvergence in $L^2$ need not follow. On $[0,1]$ with Lebesgue measure, letThen $X_n\to 0$ in probability and $\mathbb{E}|X_n{|}^2=1$, but $\Vert X_n{\Vert }_2=1$ for every $n$. This is the bounded second moments do not upgrade convergence in probability to convergence in L2 example.

Now assume $X_n\to X$ almost surely and $\mathbb{E}[X_n^2]\to \mathbb{E}[X^2]$. In particular, $(X_n)$ is bounded in $L^2$, so it is uniformly integrable in $L^1$. Almost-sure convergence and uniform integrability therefore give

We next show that $X_n$ converges weakly to $X$ in the L2 space. For $Y\in L^2$, let $Y_M=Y{1}_{\{|Y|\le M\}}$. Thenwhile the Cauchy-Schwarz inequality gives, uniformly in $n$,Letting first $n\to \infty$ and then $M\to \infty$ proves $\mathbb{E}[X_{n}Y]\to \mathbb{E}[XY]$. Taking $Y=X$ and expanding the square now givesConsequentlyThis proves the almost-sure convergence and convergence of second moments criterion.