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)


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

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    $\begingroup$ How do you quantify 'approximated by'? And yes, Lusin's theorem is the most likely candidate for homework problems of this sort. $\endgroup$ – copper.hat Jul 29 '12 at 6:49
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    $\begingroup$ It is not a homework problem ... STOP VOTING DOWN POSTS in advance - really I hate that. I was reading a paper and the authors just wrote "the usual limiting argument provides the result for all bounded functions" - prior to that they showed it for continuous functions. So I was wondering what "the usual limiting argument might be" ... Therefore I do not even know what kind of approximation is needed - uniform is alsways nice but I think pointwise would also work. $\endgroup$ – Probabilitnator Jul 29 '12 at 11:48
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    $\begingroup$ Perhaps you could post the excerpt from the original. There are several possible senses of "approximate" that could be intended here and it will be easier to give a specific answer if we know what the author is trying to accomplish. $\endgroup$ – Nate Eldredge Jul 29 '12 at 14:15
  • $\begingroup$ Okey I will try - it will be somewhat difficult, for my question emerged while reading one of the last proofs in the paper and there are lots of perliminary resuls flowing into it. $\endgroup$ – Probabilitnator Jul 29 '12 at 14:57
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    $\begingroup$ (I added my first comment under the influence, not a good idea :-).) For $1 \leq p < \infty$, $C_c(\mathbb{R}^2)$ is dense in $L^p(\mathbb{R}^2)$. (Rudin's Real & Complex Analysis", Theorem 3.14.) $C_c$ is the set of continuous functions with compact support. $\endgroup$ – copper.hat Jul 29 '12 at 16:52

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.

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 $\mu^\prime=\mu^{\prime\prime}$.

  • $\begingroup$ the Montone Class Theorem is an approach I haven't tried. I am only familiar with it from the perspective of measure theory, where one uses indicator functions and Dynkin's Theorem. Thanks $\endgroup$ – Probabilitnator Jul 29 '12 at 14:52

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 $\mu' = \mu''$.

  • $\begingroup$ I was thinking of adding this approach to my answer. You've saved me the trouble. Thanks! $\endgroup$ – user940 Jul 29 '12 at 18:20
  • $\begingroup$ Okey - one can approximate any $1_{(a,b)}$ Function via continuous functions and then use the argument you have provided to show the statement for bounded functions. Isn't that basically the monotone class theorem (the argument seems pretty familiar to me) $\endgroup$ – Probabilitnator Jul 29 '12 at 18:54
  • $\begingroup$ I assume $(a,b)$ is meant to be an open square in $\R^2$ ? $\endgroup$ – Probabilitnator Jul 30 '12 at 9:17

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 http://planetmath.org/encyclopedia/C_cXIsDenseInLpX.html.

  • $\begingroup$ So basically if $K$ is compact and I am dealing with the Borel or Lebesgue-Mesure - the approach should work? Does this theorem have a name or do you by any chance know of any citeable reference. $\endgroup$ – Probabilitnator Jul 29 '12 at 14:53
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    $\begingroup$ Most introductory texts on measure theory or functional analysis should prove this; e.g., Folland's "Real Analysis: Modern Techniques and their applications" or Royden's "Real Analysis". $\endgroup$ – user15464 Jul 29 '12 at 15:09

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