Proof of The Square of a Square-Integrable Martingale with Finite Fourth Moments is a Nonnegative Submartingale

Throughout, $0\le s\le t\le T$ and $A\in\mathcal{F}_s$ are fixed, and $\mathcal{G}$-measurability of a real-valued function on $\Omega$, for a sub-$\sigma$-algebra $\mathcal{G}$ of $\mathcal{F}$, means measurability with respect to $\mathcal{G}$ and the Borel $\sigma$-algebra of the real line.

Step 1: adaptedness, nonnegativity, and square-integrability of $M^2$. By condition (i) of the martingale definition, $M_t$ is $\mathcal{F}_t$-measurable, so $M_t^2=M_t\cdot M_t$ is $\mathcal{F}_t$-measurable by claim 3 of the arithmetic lemma for measurable functions applied on the measurable space $(\Omega,\mathcal{F}_t)$. Clearly $M_t^2(\omega)\ge0$ for every $\omega$, and $M_t^2$ is square-integrable by hypothesis. Hence the family $M^2$ satisfies conditions (i) and (ii) of the definition, and it remains to verify the submartingale inequality.

Step 2: the pointwise identity. For every $\omega\in\Omega$, expanding the square gives

$$M_t^2-M_s^2=(M_t-M_s)^2+2M_s(M_t-M_s).$$

Step 3: integrability. Write $\mathbf{1}_{A}$ for the indicator of $A$; it is an $\mathcal{F}_s$-measurable random variable (claim 1 of the arithmetic lemma) with $\mathbf{1}_{A}^2=\mathbf{1}_{A}$, hence square-integrable. By the closure properties recorded in that definition: $M_t^2\mathbf{1}_{A}$ and $M_s^2\mathbf{1}_{A}$ are integrable, as products of two square-integrable random variables; the random variable $M_tM_s$ is square-integrable, since $(M_tM_s)^2=M_t^2M_s^2\le\tfrac12\big(M_t^4+M_s^4\big)$ pointwise and the right side is integrable, so $(M_t-M_s)^2=M_t^2-2M_tM_s+M_s^2$ is square-integrable and $(M_t-M_s)^2\mathbf{1}_{A}$ is integrable; and $Z=M_s\mathbf{1}_{A}$ is square-integrable because $Z^2\le M_s^2$ pointwise, so that $Z(M_t-M_s)=M_s(M_t-M_s)\mathbf{1}_{A}$ is integrable, $M_t-M_s$ being square-integrable.

Step 4: orthogonality of the increment. The random variable $Z=M_s\mathbf{1}_{A}$ is $\mathcal{F}_s$-measurable (product of $\mathcal{F}_s$-measurable functions, claim 3 of the arithmetic lemma) and square-integrable. By condition (iii) of the martingale definition, $M_s$ is a conditional expectation of $M_t$ given $\mathcal{F}_s$, so it has the averaging property 3 of the existence and uniqueness theorem for conditional expectation; by the equivalence of properties 1--3 asserted there it also has the orthogonality property 2, which applied to $Z$ gives

$$\mathbb{E}\big[(M_t-M_s)Z\big]=0 .$$

Step 5: conclusion. Multiply the identity of Step 2 by $\mathbf{1}_{A}$ and take expectations; every term is integrable by Step 3, so by linearity of the integral,

$$\mathbb{E}\big[M_t^2\mathbf{1}_{A}\big]-\mathbb{E}\big[M_s^2\mathbf{1}_{A}\big]=\mathbb{E}\big[(M_t-M_s)^2\mathbf{1}_{A}\big]+2\,\mathbb{E}\big[(M_t-M_s)Z\big]=\mathbb{E}\big[(M_t-M_s)^2\mathbf{1}_{A}\big]\ge0,$$

the final inequality by monotonicity of the integral, the integrand being nonnegative. This is the displayed identity of the lemma, and the inequality $\mathbb{E}[M_t^2\mathbf{1}_{A}]\ge\mathbb{E}[M_s^2\mathbf{1}_{A}]$ for all $0\le s\le t\le T$ and $A\in\mathcal{F}_s$ is exactly the submartingale requirement of the definition. $\blacksquare$