I will give solution of Problem 4. I use andrewkirks notation:
andrewkirk said:
Is the following then an accurate representation of what the set ##\mathcal B## is intended to be in problem 4?
Let ##C^0## be the set of all continuous functions from ##\mathbb R\to \mathbb R##.
Given a set ##S## of functions ##\mathbb R\to\mathbb R##, we say ##S## is 'pointwise closed' if, for any countable subset ##(f_n)_{n\in\mathbb N}## of ##S## that converges pointwise to a function ##f:\mathbb R\to\mathbb R##, the function ##f## is also in ##S##.
Then ##\mathcal B## is defined to be the intersection of all sets of functions ##S## that contain ##C^0## and are pointwise closed.
micromass says that this is correct.
Here is my solution:
Since all continuous functions are Borel measurable, and all pointwise limits of Borel measurable functions are Borel measurable (W. Rudin,
Real and Complex Analysis, 3rd ed
, Corollary (a), p. 15, McGraw-Hill, Singapore, 1986), it follows that all functions in ##\mathcal B## are Borel measurable functions.
Conversely, let ##\Omega## be the set of all countable ordinals.
We define ##C^\alpha##, for ##\alpha\in \Omega##, by recursion thus:
##C^0## is as above. For ##\alpha>0##, ##C^\alpha## is the set of all pointwise limits of functions in ##\cup_{\beta<\alpha} C^\beta##.
We prove that ##\mathcal B=\cup_{\alpha\in\Omega}C^\alpha##.
To see this, we first notice that by transfinite induction, any pointwise closed set ##S## of real valued functions on ##\mathbb R## which contains ##C^0## contains ##C^\alpha## for all ##\alpha\in\Omega##. This means that ##\cup_{\alpha\in\Omega}C^\alpha\subset\mathcal B##.
Conversely, ##C^0\subset\cup_{\alpha\in\Omega}C^\alpha##. Also, for any sequence ##(f_n)_{n\in\mathbb N}## of functions in ##\cup_{\alpha\in\Omega}C^\alpha##, there is a sequence of countable ordinals ##(\alpha_n)_{n\in\mathbb N}##, such that ##f_n\in C^{\alpha_n}## for all ##n\in\mathbb N##.
The supremum of any countable subset of ##\Omega## is a countable ordinal (a consequence of ##\aleph_0^2=\aleph_0##), so there is an ##\alpha\in \Omega## such that ##f_n\in C^\alpha## for all ##n\in\mathbb N##. Therefore, if such a sequence ##(f_n)_{n\in\mathbb N}## is pointwise convergent, its limit lies in ##C^{\alpha+1}##. It follows that ##\cup_{\alpha\in\Omega}C^\alpha## is pointwise closed.
This means that ##\mathcal B\subset\cup_{\alpha\in\Omega}C^\alpha##.
Hence, ##\mathcal B=\cup_{\alpha\in\Omega}C^\alpha##.
Next, we prove by transfinite induction that if ##f,g\in C^\alpha##, then ##f+g\in C^\alpha## and ##\min(f,g)\mathcal \in C^\alpha## for all ##\alpha\in \Omega##: This is true for ##\alpha=0##. Assume that it holds for all ##\beta<\alpha##, for some ##\alpha>0## (##\alpha\in\Omega##). Let ##f,g\in C^\alpha##. Then there are two sequences ##(f_n)_{n\in\mathbb N}## and ##(g_n)_{n\in\mathbb N}## of functions in ##\cup_{\beta<\alpha}C^\beta## which converge pointwise to ##f## and ##g##, respectively. Then, to each ##n\in \mathbb N## there are ##\gamma_n<\alpha## and ##\delta_n<\alpha## such that ##f_n\in C^{\gamma_n}## and ##g_n\in C^{\delta_n}##. With ##\beta_n=\max(\gamma_n,\delta_n)##, ##f_n,g_n\in C^{\beta_n}##, and, by the induction hypothesis, ##f_n+g_n\in C^{\beta_n}## and ##\min(f_n,g_n)\in C^{\beta_n}##. Since ##(f_n+g_n)_{n\in \mathbb N}## and ##\min (f_n,g_n)_{n\in \mathbb N}## converge pointwise to ##f+g## and ##\min(f,g)##, respectively, those limit functions lie in ##C^\alpha##.
The conclusion now follows by transfinite induction.
It follows that if ##f,g\in \mathcal B##, then ##f+g\in \mathcal B## and ##\min(f,g)\in\mathcal B##.
By a similar, somewhat simpler, argument, ##f \in \mathcal B## implies ##rf\in \mathcal B## for all ##r\in \mathbb R##.
Next, let ##\mathcal E## be the family of all sets ##E\subset\mathbb R## such that ##\chi_E\in\mathcal B##, where ##\chi_E## is the
characteristic function of ##E##.
Then, ##\mathcal E## contains all bounded open intervals ##(a,b)\subset \mathbb R##, for if we put
f_n(x)=\begin{cases} 0, \qquad x\le a\\ \frac{2n}{b-a}(x-a), \qquad a<x<a+\frac{b-a}{2n}\\1,\qquad a+\frac{b-a}{2n}\le x\le b -\frac{b-a}{2n}\\<br />
\frac{2n}{b-a}(b-x),\qquad b-\frac{b-a}{2n}<x<b\\0,\qquad x\ge b\end{cases}
for all ##n\in \mathbb N##, then ##(f_n)_{n\in \mathbb N}## is a sequence in ##C^0## which converges pointwise to ##\chi_{(a,b)}##.
If ##E,F\in\mathcal E##, then, by the above, ##\chi_{\mathbb R\setminus E}=1-\chi_E\in\mathcal B## and ##\chi_{E\cup F}=\min(\chi_E+\chi_F,1)\in \mathcal B##, so ##\mathbb R\setminus E\in \mathcal E## and ##E\cup F\in\mathcal E##.
Assume that ##E_n\in\mathcal E##, for all ##n \in \mathbb N##. Then, by induction, ##\cup_{k=1}^n E_k\in\mathcal E##, for all ##n\in \mathbb N##.
It follows that ##\chi_{\cup_{n=1}^{\infty} E_n}=\lim_{n\to\infty} \chi_{\cup_{k=1}^{\infty} E_k}## (pointwise), so ##\chi_{\cup_{n=1}^{\infty} E_n}\in \mathcal B## and ##\cup_{n=1}^{\infty} E_n\in \mathcal E##.
This means that ##\mathcal E## is a ##\sigma##-algebra which contains all bounded open intervals in ##\Bbb R##. It follows that ##\mathcal E## contains all Borel sets.
From this and the above, it follows that ##\mathcal B## contains all
simple Borel measurable functions, since these are linear combinations of characteristic functions of Borel sets.
Now, if ##f:\mathbb R\to [0,\infty)## is a
positive Borel measurable function, then there is a sequence ##(s_n)_{n\in\mathbb N}## of simple Borel meaurable functions which converges pointwise to ##f## (W. Rudin,
Real and Complex Analysis, 3rd ed
, Theorem 1.17, p. 15, McGraw-Hill, Singapore, 1986). Hence ##f\in \mathcal B##. Every real Borel measurable function ##f## can be written as a difference ##f_+-f_-##, where
##f_+(x)=\begin{cases}f(x),\qquad f(x)\ge 0\\ 0,\qquad f(x)<0\end{cases}## and ##f_-=f-f_+##, and ##f_+## and ##f_-## both are positive Borel measurable functions, so ##f\in \mathcal B##.
It follows that ##\mathcal B## contains all Borel measurable functions.
We have proved that ##\mathcal B## is the set of Borel measurable functions.