Krdytkyu

Let $X$ be an uncountable set, let $frak{M}$ be the collection of all sets $E subseteq X$ such that either $E$ is countable or $E^c$ is countable, and define $mu (E) = 0$ in the first case, $mu (E) = 1$ in the second case. Porve that $frak{M}$ is a $sigma$-algebra in $X$ and that $mu$ is a measure on $frak{M}$. Describe the corresponding measurable functions and their integrals.

I have already worked on the first part, but I am having trouble describing the measurable functions and their integrals, so I consulted this solution, but I am having trouble following it(see page 3). Here is the relevant passage:

The measurable functions on $frak{M}$ consist of those functions $f : X to Bbb{R}$ such that for each $r in Bbb{R}$, $f^{-1}(r)$ is at most countable or $f^{-1}(Bbb{R}- {r})$ is at most countable. If we let $A subseteq Bbb{R}$ denote the set of points such that $f^{-1}(r)$ is not countable, then the integral of $f$ is $sum_{r in A} r$.

First let's compute the integral of a simple function $s(x) = sum_{i=1}^n a_i 1_{A_i}(x)$, where the $A_i$ are pairwise disjoint, measurable sets and $X = bigcup_{i=1}^n A_i$. Then $A_k$ must be uncountable and therefore $A_k^c$ is countable. Then $A_i cap A_k = emptyset$ for every $i neq k$ implies $A_i^c cup A_k^c = X$ for $i neq k$ and therefore $A_i$ is uncountable for every $ineq k$. Hence $mu(A_i) = 0$ and therefore $mu(X cap A_i) = 0$ for $ineq k$, and $mu(X cap A_k)=0$. Hence, $displaystyle int_X s d mu = sum_{i=1}^n a_i mu(X cap A_i)= a_k$.

Now, if $f$ is measurable, then I agree with the above quote that either $f^{-1}(r)$ or $f^{-1}(Bbb{R}-{r})$ is countable for every $r in Bbb{R}$. However, I don't see how

$$int_X f d mu = sup { int s d mu mid s text{ simple, } 0 le s le f }$$

$$= sup { a in Bbb{R} mid 0 le a le f(x)~ forall x in x }$$

equals the sum $sum_{r in A} r$.

Also, I found this MSE post, but I don't quite understand Daniel Robert-Nicoud answer, particularly the second point he makes which is

Assume there is no such $x$. Then $f$ defines a partition of $X$ into at most countable sets by $bigsqcup_{xinmathbb{R}}E_x = X$. By cardinality arguments, there must be an uncountable number of sets in the partition. In particular, $f(X)$ is an uncountable subset of $mathbb{R}$, and as such it has uncountably many limit points. From this, you should be able to prove that such a function cannot exist (else you would be able to construct two disjoint uncountable subsets of $X$ that are both in $mathfrak{M}$).

What exactly are these "cardinality arguments" leading to the conclusion "there must be an uncountable number of sets in the partition"? And how does this "in particular" imply $f(X)$ is uncountable? The $E_x$ are preimages, while $f(X)$ is an image...

EDIT:

Here is something I tried. Note that $f^{-1}([x,x+1))$ is measurable for every $x in Bbb{Z}$, and $X = bigcup_{x in Bbb{Z}} f^{-1}([x,x+1))$ which means that $f^{-1}([x,x+1))$ is uncountable for some integer $x$. This means that its complement is countable and therefore $mu(f^{-1}([x,x+1)) = 1$. Since the complement is countable, $f^{-1}([x,x+1)) cap f^{-1}([y,y+1)) = emptyset$, where $x= neq u$, implies $f^{-1}([x,x+1)^c) cup f^{-1}([y,y+1)^c)$ = X which implies $f^{-1}([y,y+1)^c)$ is uncountable and therefore $f^{-1}([y,y+1))$ is countable, which means $mu (f^{-1}([y,y+1)) =0$. Hence, the integral of $f$ is (I believe):

$$int_X f d mu = int_{bigcup f^{-1}([z,z+1))} f d mu = sum_{z in Bbb{Z}} int_{f^{-1}([z,z+1))} f d mu = int_{f^{-1}([x,x+1))} f d mu $$

Of course, assuming that is right, I still need to evaluate $int_{f^{-1}([x,x+1))} f d mu$ which I am unable to do at the moment.

Part I: Typos and mistakes in your attempts

Let me first point out it appears you have a typo in the sentence

Then $A_i cap A_k = emptyset$ for every $i neq k$ implies $A_i^c cup A_k^c = X$ for $i neq k$ and therefore $A_i$ is uncountable for every $i neq k$.

You must have meant to say

. . . and therefore $A_i$ is countable for every $i neq k$.

Secondly, there is no need to write that the integral of the simple function $s(x) = sum_{i=1}^n a_i 1_{A_i}(x)$ over $X$ w.r.t. the measure $mu$ is
$$
int_X s, dmu = sum_{i=1}^n a_i mu(A_i cap X);
$$
you can simply write $mu(A_i)$ in place of $mu(A_i cap X)$ since $A_i cap X = A_i$ for all $1 leq i leq n$. Only when you are integrating over a proper subset of $X$, say $E$, do you need to carefully specify that the integral is $sum_{i=1}^n a_i mu(A_i cap E)$.

Thirdly, the set
$$
sup { a in Bbb{R} mid 0 leq a leq f(x) forall x in X } qquad (text{note the typo: you have } x in x)
$$
is not clearly expressed. Is presume that $a$ is meant to be independent of $x$, in which case this set is just $inf{ f(x) : x in X }$, and we definitely don't have equality between this set and
$$
sup { int s, dmu mid s text{ simple, } 0 leq s leq f }.
$$

Lastly, in your edit, the following part is completely unclear to me.

Since the complement is countable, $f^{-1}([x,x+1)) cap f^{-1}([y,y+1)) = emptyset$, where $x= neq u$, implies $f^{-1}([x,x+1)^c) cup f^{-1}([y,y+1)^c)$ = X which implies $f^{-1}([y,y+1)^c)$ is uncountable and therefore $f^{-1}([y,y+1))$ is countable, which means $mu (f^{-1}([y,y+1)) =0$.

Perhaps the argument can be fixed, but as it stands it does not make sense. For instance, it is false that $f^{-1}([x,x+1)) cap f^{-1}([y,y+1)) = emptyset$.

Part II: About your questions on @DanielRobert-Nicoud's answer

We assume that $f : X to Bbb{R}$ is measurable and for each $x in Bbb{R}$ we define $E_x in mathfrak{M}$ by $E_x = f^{-1}(x)$. Suppose there is no $x in Bbb{R}$ such that $E_x$ is uncountable. Then, we have $X = bigsqcup_{x in Bbb{R}} E_x$ with each $E_x$ being at most countable.

Note that many of the sets $E_x$ can be empty. If there were at most countably many $x in Bbb{R}$ such that $E_x neq emptyset$, then $bigsqcup_{x in Bbb{R}} E_x$ would be countable, which is a contradiction. These are the "cardinality arguments" that show that there must be uncountably many nonempty sets in this partition of $X$.

In fact, let us examine this more carefully. For each point $x$ in the image $f(X) subset Bbb{R}$, we have a distinct nonempty measurable set $E_x subset X$. What we have just concluded is that there are uncountably many $x in Bbb{R}$ such that $E_x$ is nonempty. Moreover, $E_x$ is nonempty only if $x$ lies in the image of $f$, by definition of $E_x$. Both facts together can be summarised by saying, we have uncountably many $x in f(X)$. This is the claim that was made after saying, "In particular".

Part III: My solution

I will note that the first solution you have quoted from an online resource is unsatisfactory. It hardly tells us anything more than the definition of a measurable function in this particular case. Same for the description of the integral.

@DanielRobert-Nicoud's answer is accurate. I would like to give a proof in my own words though, just for my satisfaction.

Since any measurable $f : X to Bbb{R}$ can be written as $f = f^+ - f^-$, let us first try to describe $f$ when it is positive, that is its range lies in $[0,infty)$. In this case, we know that $f$ is the pointwise limit of a monotonically increasing sequence of positive simple functions. So, let us take $s : X to Bbb{R}$ to be a positive simple function. We can write
$$
s = sum_{i=1}^n alpha_i chi_{A_i}, qquad A_i in mathfrak{M},alpha_i geq 0.
$$
We assume that the $A_i$ are pairwise disjoint. So, precisely one $A_i$ has at most countable complement, while the rest are themselves at most countable.

Now, let $s_n : X to Bbb{R}$ be a simple function for each $n in Bbb{N}$ such that $0 leq s_1 leq s_2 leq dots leq f$ and $lim_{n to infty} s_n(x) = f(x)$ for all $x in X$. To each $s_n$ we associate the unique measurable set of at most countable complement that we discussed above, and we shall denote it by $E_n$. Let $alpha_n = s_n(x)$ for any $x in E_n$. Then, we have that
$$
0 leq alpha_1 leq alpha_2 leq dots leq f(x) quad text{ for all }quad x in E := bigcap_{n=1}^infty E_n,
$$
and $lim_{n to infty} alpha_n = f(x)$ for all $x in E$. Note that $E$ has at most countable complement. The same observation holds for an arbitrary measurable function $f$ (that is, not necessarily positive measurable function). Hence, if $f : X to Bbb{R}$ is a measurable function, $f$ must be constant on an uncountable measurable set.

The converse is also true. If $f$ is constant on $E$, an uncountable measurable set, then for any definition of $f(x)$ for the points $x in X - E$, we can show that $f$ turns out to be measurable. Indeed, since $X - E$ is countable, it has an enumeration, say ${ x_n }_{n in Bbb{N}}$. Assume that $f$ is a positive measurable function. Suppose we define $f(x_n) = alpha_n geq 0$ and $f(x) = alpha geq 0$ for $x in E$. Then, we define $s_n : X to Bbb{R}$ by
$$
s_n = alpha chi_E + sum_{i=1}^n alpha_i chi_{{x_i}}.
$$
It is easy to see that $0 leq s_1 leq s_2 leq dots leq f$ and $lim_{n to infty} s_n(x) = f(x)$ for all $x in X$.

With this description of $f$, the integral of $f$ becomes simple to evaluate: it is just $alpha$, the constant value that $f$ takes on an uncountable measurable set.

Note that it is still true that the integral of $f$ is
$$
sum_{r in A} r,
$$
where $A subset Bbb{R}$ is the set of points $r in Bbb{R}$ such that $f^{-1}(r)$ is uncountable. It is just that $A$ is the singleton set ${ alpha }$, so the summation is quite trivial.