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

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

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,

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. $\blacksquare$