This is not possible. Take $(\Omega_i,\mathcal{A}_i)=([0,1],\mathcal{B})$ to be the unit interval with the Borel $\sigma$-algebra and $\mathcal{M}_i=\mathcal{B}$ for $i=1,2$. Let $H$ be the indicator function of the diagonal, certainly a measurable function. Let $H_n$ be a linear combination of measurable rectangles such that $0\leq H_n\leq H$. Since every measurable rectangle that is a subset of the diagonal must be a singleton of the form $\{(x,x)\}$, each $H_n$ has only finitely many nonzero values. Taking the union of these values over all $H_n$, we get a countable set of such values. But since the diagonal is uncountable, we cannot get convergence at every point.

$\mathcal{H}$ does not contain every measurable function. Take $(\Omega_i,\mathcal{A}_i)=([0,1],\mathcal{B})$ to be the unit interval with the Borel $\sigma$-algebra and $\mathcal{M}_i=\mathcal{B}$ for $i=1,2$. 

Let $H$ be the indicator function of the diagonal, certainly a measurable function. Let $H_n$ be a linear combination of measurable rectangles such that $0\leq H_n\leq H$. That $0\leq H_n$ is without loss of generality since for every linear combination $G$ of measurable rectangles, $G\vee 0$ is still a linear combination of measurable rectangles.

Since every measurable rectangle that is a subset of the diagonal must be a singleton of the form $\{(x,x)\}$, each $H_n$ has only finitely many nonzero values. Taking the union of these values over all $H_n$, we get a countable set of such values. But since the diagonal is uncountable, we cannot get convergence at every point.