Approximation of bounded measurable functions with continuous functions

Question

This is not homework. I was reading a paper where the authors showed a result for all continuous functions and then just proceeded to write "the usual limiting Argument gives the result for all bounded functions" - so I am asking myself what this "usual limiting argument" might be. I do not know whether they mean uniform or pointwise convergence. As I see it pointwise convergence should suffice :D

Thus I was am wondering whether there is a theorem having or leading to the following statement:

Let $K\subset\mathbb{R}^2$ be compact. Any bounded measurable function $f:K\to\mathbb{R} $ can be approximated by a sequence of continuous functions $(g_m)$ on $K$.

Nate Eldredge suggested that I post some excerpt from the original to provide more context for the problem. Here I go:

The goal is to proof the existence of a weak limit for a tight sequence of probability measures on $\mathcal{C}^0([0,1]^2,\mathbb{R})$ associated with reflecting Brownian Motions on the compact the set $[0,1]^2$ which is a Lipshitz Domain. Thus we already, know that some weak limit must exist and it remains to show that two limit-Points agree. Weak-Convergence is generally defined via bounded measurable functions. Now let $P'$ and $P''$ be two subsequential limit points. The authors show that $f \in \mathcal{C}^0([0,1]^2,\mathbb{R})$ the following holds (here $X_s$ the canonical process)

$E'f(X_s)=E''f(X_s)$

And now comes the actual source of my question: "The usual limiting argument gives the result for bounded $f$ and hence the one-dimensional distributions agree." (the second part I understand only the "standard limiting argument thing" is somewhat confusing)

Any Help is much appreciated and Thanks in Advance :D

Answer 1

To respond to your comment on Byron's answer:

The functional monotone class theorem is a very useful result and well worth knowing. However, you can also get this result with arguments that may be more familiar. To recap, we want to show:

Suppose $\mu', \mu''$ are two probability measures on $\mathbb{R}$, and we have $\int f\,d\mu' = \int f\,d\mu''$ for all bounded continuous $f$. Then $\mu' = \mu''$.

One could proceed as follows:

Exercise. For any open interval $(a,b)$, there is a sequence of nonnegative bounded continuous functions $f_n$ such that $f_n \uparrow 1_{(a,b)}$ pointwise.

(For example, some trapezoidal-shaped functions would work.)

If $f_n$ is such a sequence, we have $\int f_n \,d\mu' = \int f_n \,d\mu''$ for each $n$. By monotone convergence, the left side converges to $\int 1_{(a,b)}\,d\mu' = \mu'((a,b))$ and the right side converges to $\mu''((a,b))$. So $\mu'((a,b)) = \mu''((a,b))$, and this holds for any interval $(a,b)$.

Now you can use Dynkin's $\pi$-$\lambda$ lemma, once you show:

Exercise. The collection $$\mathcal{L} := \{B \in \mathcal{B}_\mathbb{R} : \mu'(B) = \mu''(B)\}$$ is a $\lambda$-system. (Here $\mathcal{B}_{\mathbb{R}}$ is the Borel $\sigma$-algebra on $\mathbb{R}$.)

We just showed that the open intervals are contained in $\mathcal{L}$. But the open intervals are a $\pi$-system which generates $\mathcal{B}_{\mathbb{R}}$. So by Dynkin's lemma, $\mathcal{B}_\mathbb{R} \subset \mathcal{L}$, which is to say $\mu' = \mu''$.

Answer 2

It is not true that every bounded measurable function is the pointwise, or uniform, limit of continuous functions. See this MSE question.

It depends on what result the author wanted to prove, but extending results from continuous functions to bounded measurable functions often uses the Monotone Class Theorem.

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Added: You have two probability measures $\mu^\prime:=P^\prime\circ X_s^{-1}$ and $\mu^{\prime\prime}:=P^{\prime\prime}\circ X_s^{-1}$ on $[0,1]^2$ so that, for every continuous function $f$, $$\int f\,d\mu^\prime=\int f\,d\mu^{\prime\prime}.\tag1$$

Let $\cal H$ be the space of all bounded, measurable functions $f$ so that (1) holds, and let $\cal K$ be the space of continuous functions. Once you check the conditions of the functional Monotone Class Theorem, you can conclude that $\cal H$ contains all bounded functions measurable with respect to $\sigma(\cal K)$. That is, $\cal H$ includes all bounded, Borel measurable functions which means that $\mu^\prime=\mu^{\prime\prime}$.

Answer 3

Most likely this was referring to approximation in the $L^p$ norm ($p\geq 1$),

$$\|f\|_p = \left(\int |f|^p d\mu \right)^{1/p}.$$

It is true that if $\mu$ is sufficiently well behaved, then every function $f$ such that $\|f\|_p$ is finite (i.e. $f\in L^p$) can be approximated in this norm by continuous functions. A proof can be found here.