Proof of The Gradient of a Convex Continuously Differentiable Function Belongs to the Tangent Space When It Is Square-Integrable

Each result cited is universally quantified over the data in its own statement.

Borel measurability. The components of $\nabla f$ are the partial derivatives $\partial_{i}f$, $i\in[d]$, which are continuous on $\mathbb{R}^{d}$ because $f$ is of class $C^{1}$ (Differential Calculus and Convexity on Euclidean Open Sets: Standing Notation §derivatives); a map into $\mathbb{R}^{d}$ with continuous components is Borel by Probability Measures on Euclidean Space and Random Vectors: Standing Notation §borel-maps.

The subdifferential is the gradient. Let $x\in\mathbb{R}^{d}$. By A Real-Valued C^1 Function is Differentiable at Every Point, applied with $U=\mathbb{R}^{d}$ and $a=x$, the function $f$ is differentiable at $x$ with derivative matrix the real matrix with one row and $d$ columns whose entry in column $i$ is $\partial_{i}f(x)$. The point of $\mathbb{R}^{d}$ whose $i$th coordinate is that entry is $Df(x)=\nabla f(x)$. Since $\mathbb{R}^{d}$ is open and convex (as recorded in the statement) and $f$ is convex on it, Elementary Calculus of the Subdifferential of a Convex Function §gradient gives

with $\partial_{\mathbb{R}^{d}}f$ the subdifferential.

Tangency. Suppose $\int_{\mathbb{R}^{d}}\lVert\nabla f\rVert^{2}\,d\mu<\infty$. Apply A Square-Integrable Selection of the Subdifferential of a Convex Potential Belongs to the Tangent Space §tangent with $G=\mathbb{R}^{d}$ (open and convex), $\phi=f$, $D=\mathbb{R}^{d}$ (a Borel subset of $G$ with $\mu(D)=1$, $\mu$ being a probability measure) and $T=\nabla f$, which is Borel, satisfies $\partial_{G}\phi(x)=\{T(x)\}$ for every $x\in D$ by the previous step, and is square-integrable by assumption. It gives $\nabla f\in T_{\mu}$. $\blacksquare$