Suppose that $(X,d)$ is a locally compact metric space and $\mu$ is a $\sigma$-finite Radon measure on the Borel sigma-algebra of this space. I am aware that if $(X,d)$ is separable and $\mu$ has full support then $L^2(X,\mu)$ is separable i.e. it admits a dense countable subset.


My question is the following: does there exist a *countable* subset $G \subset L^2(X,\mu) \cap C\_c(X)$ such that



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> for every $f \in C\_c(X) \cap L^2(X,\mu) \cap L^\infty(X,\mu), f \geq 0$ a.s., there exists an *increasing* (i.e. $f\_n \geq f\_m$ for all $n \geq m$) sequence of functions $f\_n \in G$ such that $f\_n \uparrow f$ pointwise?
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> 
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If this occurs, then by the monotone convergence theorem, we will have $f\_n \to f$ in $L^2$ as well, which is great.


Essentially, given the existence of such a countable $G$ allows me to do the following : given $f \in C\_c(X)$ and such a sequence, I can obtain an inequality of the form $L(f\_i-f\_j)(x) \leq C\|f\_i-f\_j\|\_{2}^2$, however this can only be obtained when $f\_i-f\_j \geq 0$ a.e. This becomes an inequality of the form $L(f\_i)(x)-L(f\_j)(x) \leq C\|f\_i-f\_j\|^2\_2$ as $L$ is linear, and because the RHS converges to $0$ as $i,j \to \infty$, $L(f\_i)(x) \to Lf(x)$ (the argument goes on from here, but that's basically why non-negativity is required).


Note : I asked a slightly similar question over at [Mathematics Stack Exchange](https://math.stackexchange.com/questions/4636432/density-of-functions-in-increasing-pointwise-sense) as well, it's a simplification of this question that can be attempted before this one.



On the existence of a countable dense family in "increasing" pointwise convergence
Yes, this is true. First note that if there is a sequence $f\_n$ that converges to $f$ from below, $f\_n \leq f$, then there is an increasing sequence converging to $f$ given by $\tilde{f}\_n = max(f\_1,...,f\_n)$.
You can add such functions to your dense countable set without changing its cardinal.


So we are left to show that we can find a sequence converging from below.
Let me give the argument when $X = {\bf R}$ for simplicity. You can find a simple function less than $f$ whose integral is $\varepsilon$-close to the integral of $f$, from a set of countable simple functions, the one with rational values and associated to intervals with rational endpoints. This is actually true for any Riemann-integrable functions. This simple function can be approximated from below by an affine function, here again from a countable subset of the sets of all affine functions, e.g. the one with rational slope and rational turning points.


The same process work for general separable locally compact metric spaces, you just need to check how the approximating functions from the countable set mentioned in your question are built.


