# A question concerning the family of bounded Borel functions

This is the question 16, chapter 1 in "Functional Analysis" by Reed and Simon. I've already encountered that the question was previously asked, but no answer that helped me was provided, hence I'm posing it here again and commenting my thoughts. I'd greatly appreciate if someone could help me out! Thank you!

The question is:

Prove that the bounded Borel functions on $[0;1]$ are the smallest family $\mathcal{F}$ which includes $C[0;1]$ and has the property: If $f_n$ is a sequence of uniformly bounded functions in $\mathcal{F}$ and $f_n \rightarrow f$ pointwise, then $f\in\mathcal{F}$.

We shall now denote this property by $(\star)$.

Now my understanding of the question is that, (1) I have to prove that $C[0;1] \subseteq \mathcal{F}$, ($\mathcal{F}$ denoting the the family of all bounded Borel functions on $[0;1]$), (2) the family $\mathcal{F}$ satisfies property $(\star)$ and (3) that for every family $\mathcal{G}$ which includes $C[0;1]$ and satisfies property $(\star)$, $\mathcal{F} \subseteq \mathcal{G}$.

Now (1) is fairly easy, as every continuous function is Borel, and any continuous function on a compact set is bounded.

(2) is straightforward as well. For $f_n$ a sequence of Borel functions, the pointwise limit $f$ is Borel as well, and similarly a pointwise limit of a sequence of uniformly bounded functions is uniformly bounded with the same bound.

Now my perspective is that statement (3) is equivalent to stating that every bounded Borel function is a pointwise limit of uniformly bounded continuous functions: As for an arbitrary family $\mathcal{G}$, we only know that it contains the continuous functions, and it is closed under pointwise limits (for sequences of uninformly bounded functions). As far as I know though, the function $\chi_{\mathbb{Q} \cap [0;1]}$ (indicator function of the rationals in the interval $[0;1]$) is Borel (for $\mathbb{Q}\cap[0;1]$ is a Borel set), clearly bounded, but not the limit of a sequence of continuous functions, as it is discontinuous everywhere (see here).

Now that means that either the question is erroneous (which I highly doubt) or that my understanding of what I have to prove is wrong.

Could someone explain to me where I made a mistake above? Thanks a lot!

Statement (3) is not equivalent to saying that every bounded Borel function is a pointwise limit of a uniformly bounded sequence of continuous functions. Rather, it is equivalent to every bounded Borel function being either (1) a pointwise limit of a uniformly bounded sequence of continuous functions, or (2) a pointwise limit of a uniformly bounded sequence of functions as in (1), or (3) a pointwise limit of a uniformly bounded sequence of functions as in (2), or ... (continued transfinitely for uncountably many steps). In other words, you have to consider limits of limits of limits of ... of continuous functions, and the iteration in "..." has to be long enough so that taking limits again gives nothing new; that "long enough" requires an uncountably long iteration.

• My question would be now, how do you show that this limiting process in fact exhausts all Borel functions? – Marcel S Mar 19 '18 at 18:00

No, (3) does not say that every bounded function is the limit of a sequence of continuous functions: Say $L$ is the set of all pointwise limits of uniformly bounded sequences of continuous functions. Then $L$ is not closed under pointwise sequential limits.

Blass has given the cool way to look at all this, in terms of transfinite induction. One might call that the bottom-up approach. As with questions about sigma algebras there's also a top-down approach.

Thanks to Nate Eldredge and John Dawkins for pointing out problems with the first few versions of the following, and for suggesting a much more elegant fix for one than the patch I came up with.

Let's say a set $S$ of functions is closed if it's closed under uniformly bounded sequential pointwise convergence (if $f_n\in S$, $f_n$ is uniformly bounded and $f_n\to f$ pointwise then $f\in S$).

So we assume that $C\subset\mathcal G$ and $\mathcal G$ is closed, and we need to show that $\mathcal G$ contains every bounded Borel function.

Let $\mathcal G_0$ be a maximal vector space such that $C\subset\mathcal G_0\subset\mathcal G$. (Zorn's lemma shows there is such a thing.)

Let $\mathcal G_1$ be the space of all limits of uniformly bounded pointwise convergent sequences of elements of $\mathcal G_0$. It's clear that $\mathcal G_1$ is a vector space and that $\mathcal G_0\subset \mathcal G_1\subset \mathcal G$, so the maximality of $\mathcal G_0$ shows that $\mathcal G_1=\mathcal G_0$. That is, $\mathcal G_0$ is closed.

Let $A$ be the algebra consisting of finite unions of sets of the form $[a,b)\subset[0,1]$ and $[a,1]\subset[0,1]$.

(i) Suppose $E\in A$. There exists a sequence $(f_n)\subset C$ with $0\le f_n\le 1$ and $f_n\to\chi_E$ pointwise; since $C\subset\mathcal G_0$ and $\mathcal G_0$ is closed it follows that $\chi_E\in \mathcal G_0$.

(ii) The fact that $\mathcal G_0$ is closed shows that the set of $E$ with $\chi_E\in\mathcal G_0$ is a monotone class.

(iii) The monotone class theorem now shows that $\chi_E\in\mathcal G_0$ for every Borel set $E$.

(iv) Since $\mathcal G_0$ is a vector space it contains every simple Borel function.

(v) Since $\mathcal G_0$ is closed it contains every bounded Borel function.

• I think step (ii) is not so easy. This really needs a monotone class type argument. – Nate Eldredge Mar 19 '18 at 18:00
• @NateEldredge I believe you're right, thanks. – David C. Ullrich Mar 19 '18 at 18:17
• How is (iv) going to go without some vector space structure for $\mathcal G$? – John Dawkins Mar 19 '18 at 18:58
• @JohnDawkins Another good question. I'm starting to wonder whether the top-down argument is as clear as I thought. – David C. Ullrich Mar 19 '18 at 19:42
• A glance at the first chapter of Dellacherie and Meyer's Probabilities and Potential suggests the following modification of your (last but one) argument: Let $\mathcal G_0$ denote a maximal min-stable vector space containing $C[0,1]$ and contained in $\mathcal G$. (Zorn's lemma!) Your steps, with $\mathcal G_0$ replacing $\mathcal G$, show that $\mathcal G_0$ contains all bounded Borel functions, hence so does $\mathcal G$. – John Dawkins Mar 19 '18 at 20:53