There are many well known results on approximations by regular functions. Here are some of them.
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If $f : \mathbb R \to \mathbb R$ is (Lebesgue) measurable, then there exists a sequence of continuous functions $f_n : \mathbb R \to \mathbb R$ such that $f_n \to f$ pointwise a.e. (see here for a proof).
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If $f \in C(\mathbb R)$ is a continuous function then there exists a sequence of functions $f_n \in C^{\infty}(\mathbb R)$ such that $f_n \to f$ uniformly on every compact subset of $\mathbb R$. Furthermore, if $f \in C^K(\mathbb R)$ then the convergence can be ensured to uniformly hold on every compact set for all derivatives up to order $K$.
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The set $C_c^{\infty}(\mathbb R^d)$ is dense in each of the spaces $L^p(\mathbb R^d), 1 \leq p < \infty$ in their norms.
I am aware that pointwise a.e. convergence is a very weak form of convergence to ask for. A tool like Lusin's theorem with an approximation argument is enough to guarantee a convergence result. Keeping this in mind, I'm thinking that a slightly stronger version of this convergence, which I require, should also be true. The key difference with this problem is that I require a uniform bound on the approximating functions in question.
Suppose that $g : \mathbb R \to \mathbb R$ is a Borel measurable function such that $|g|$ is essentially bounded above by $1$. Then, does there exist a sequence of functions $\{f_n\}_{n \geq 1}$, such that each $f_n$ is twice differentiable with bounded and continuous derivatives, $|f_n| \leq 1$ on $\mathbb R^d$ for all $n$, and $f_n \to g$ pointwise a.e.?
Now, without the condition $|f_n| \leq 1$ on $\mathbb R^d$, there is no problem at all. We may consider a sequence of continuous functions $h_n$ which approach $g$ pointwise a.e., and for each $h_n$ we can consider a sequence of smooth functions $h_{ni} \to h_n$ pointwise a.e. We'll be done by a diagonal argument. However, it's not possible to place a restriction on $|h_{ni}|$, it seems, in any particular way. That insight is what I'm looking for here.
Best Answer
The trick here is to use bump functions. Specifically, we are using the fact that if $K\subseteq U$ with $K$ compact and $U$ open, and $c_1$ and $c_2$ are constants, then there is a smooth (i.e., in $C^\infty(\mathbb R^d)$) function equaling $c_1$ on $K$ and $c_2$ outside of $U$, and with values between the two constants everywhere.
Let $h_n$ be a sequence of smooth functions converging pointwise a.e. to $g$, as you have described. Notice also that if $\psi_n$ are smooth bump functions equaling $1$ on the balls of radius $n$ about the origin, and then decaying to $0$, then $h_n\psi_n$ also converges pointwise a.e. to $g$, so henceforth we assume WLOG that each $h_n$ is compactly supported.
For each $n$, let $U_n=\{x\in\mathbb R^n\mid |h_n(x)|> 1+\frac{1}{n}\}$, and let $K_n=\{x\in\mathbb R^n\mid |h_n(x)|\geq 1+\frac{2}{n}\}$. Then $K_n\subseteq U_n$ with $K_n$ compact and $U_n$ open, so let $\phi_n$ be a smooth bump function bounded by $\frac{1}{1+\frac{2}{n}}$, equal to $\frac{1}{1+\frac{2}{n}}$ outside of $U_n$ and equal to $0$ on $K_n$. Then each $h_n\phi_n$ is smooth and compactly supported (hence all derivatives bounded) and bounded by $1$ in absolute value everywhere, and $h_n\phi_n\rightarrow g$ pointwise a.e. (since at any $x$ where $h_n(x)$ converges to $g(x)$, eventually $x$ will lie outside of every $U_n$).