[proofplan] Adaptedness and integrability are already part of the hypotheses, so only the [conditional expectation](/page/Conditional%20Expectation) identity remains. For fixed $0 \leq s \leq t$, write $X_t$ as the sum of the past value $X_s$ and the future increment $X_t-X_s$. The past value is $\mathcal F_s$-measurable and can be pulled through conditional expectation, while the increment is independent of $\mathcal F_s$ and centered, so its conditional expectation is zero. [/proofplan] [step:Reduce the martingale property to the conditional expectation identity] By hypothesis, for every $t \geq 0$, the [random variable](/page/Random%20Variable) $X_t:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is integrable and $\mathcal F_t$-measurable. Thus $(X_t)_{t \geq 0}$ is adapted and integrable. It remains to prove that for every $0 \leq s \leq t$, $\mathbb E[X_t\mid \mathcal F_s]=X_s$ a.s. Fix $0 \leq s \leq t$. Since $X_s$ and $X_t$ are integrable, the increment $Y: \Omega \to \mathbb R$, $\omega \mapsto X_t(\omega)-X_s(\omega)$, is integrable. [/step] [step:Show that the centered independent increment has zero conditional expectation] We claim that \begin{align*} \mathbb E[Y\mid \mathcal F_s]=0 \end{align*} a.s. Let $\sigma(Y)$ denote the smallest $\sigma$-algebra on $\Omega$ with respect to which the random variable $Y:(\Omega,\mathcal F)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable. The phrase "$Y$ is independent of $\mathcal F_s$" means that the $\sigma$-algebras $\sigma(Y)$ and $\mathcal F_s$ are independent. Let $A \in \mathcal F_s$. Since $Y$ is independent of $\mathcal F_s$, the measures \begin{align*} \mu_A(B):=\mathbb P(A\cap \{Y\in B\}) \end{align*} and \begin{align*} \nu_A(B):=\mathbb P(A)\mathbb P(Y\in B) \end{align*} on $(\mathbb R,\mathcal B(\mathbb R))$ agree for every $B\in\mathcal B(\mathbb R)$. Hence, by equality of integrals with respect to equal finite measures and by integrability of $Y$, \begin{align*} \int_A Y\,d\mathbb P=\mathbb P(A)\int_\Omega Y\,d\mathbb P. \end{align*} The centering hypothesis gives $\int_\Omega Y\,d\mathbb P=\mathbb E[Y]=0$, so \begin{align*} \int_A Y\,d\mathbb P=0. \end{align*} Since this holds for every $A\in\mathcal F_s$, the defining property and uniqueness of conditional expectation give \begin{align*} \mathbb E[Y\mid \mathcal F_s]=0 \end{align*} a.s. [guided] We need to justify the standard step that an independent centered increment has conditional expectation zero. Define the increment as the integrable random variable \begin{align*} Y: \Omega \to \mathbb R, \qquad \omega \mapsto X_t(\omega)-X_s(\omega). \end{align*} Let $\sigma(Y)$ denote the smallest $\sigma$-algebra on $\Omega$ with respect to which $Y$ is measurable. The hypothesis that $Y$ is independent of $\mathcal F_s$ means that $\sigma(Y)$ and $\mathcal F_s$ are independent: for every $A\in\mathcal F_s$ and every $B\in\mathcal B(\mathbb R)$, \begin{align*} \mathbb P(A\cap \{Y\in B\})=\mathbb P(A)\mathbb P(Y\in B). \end{align*} Fix an event $A\in\mathcal F_s$. Define two finite measures on $(\mathbb R,\mathcal B(\mathbb R))$ by \begin{align*} \mu_A(B):=\mathbb P(A\cap \{Y\in B\}) \end{align*} and \begin{align*} \nu_A(B):=\mathbb P(A)\mathbb P(Y\in B). \end{align*} The independence identity says exactly that $\mu_A(B)=\nu_A(B)$ for every Borel set $B$. Therefore the two measures are equal. Since $Y$ is integrable, the identity function on $\mathbb R$ is integrable with respect to these pushforward-type finite measures, and equality of the measures gives \begin{align*} \int_A Y\,d\mathbb P=\int_{\mathbb R} x\,d\mu_A(x)=\int_{\mathbb R} x\,d\nu_A(x)=\mathbb P(A)\int_\Omega Y\,d\mathbb P. \end{align*} Now the increment is centered by hypothesis, so \begin{align*} \int_\Omega Y\,d\mathbb P=\mathbb E[Y]=0. \end{align*} Thus for every $A\in\mathcal F_s$, \begin{align*} \int_A Y\,d\mathbb P=0. \end{align*} The zero random variable is $\mathcal F_s$-measurable and integrable, and it has the same integrals over all events $A\in\mathcal F_s$ as $Y$. By the defining uniqueness property of conditional expectation, this proves \begin{align*} \mathbb E[Y\mid \mathcal F_s]=0 \end{align*} a.s. [/guided] [/step] [step:Pull out the past value and conclude the martingale identity] Because $(X_t)_{t\geq 0}$ is adapted, $X_s$ is $\mathcal F_s$-measurable. Since $X_s$ and $Y$ are integrable, linearity of conditional expectation and the pull-out property for $\mathcal F_s$-measurable integrable random variables give \begin{align*} \mathbb E[X_t\mid \mathcal F_s]=\mathbb E[X_s+Y\mid \mathcal F_s]=X_s+\mathbb E[Y\mid \mathcal F_s]. \end{align*} By the previous step, $\mathbb E[Y\mid \mathcal F_s]=0$ a.s., hence \begin{align*} \mathbb E[X_t\mid \mathcal F_s]=X_s \end{align*} a.s. Since $0\leq s\leq t$ was arbitrary, $(X_t)_{t\geq 0}$ satisfies the defining conditional expectation identity for a martingale. Together with adaptedness and integrability, this proves that $(X_t)_{t\geq 0}$ is a martingale with respect to $(\mathcal F_t)_{t\geq 0}$. [/step]