Real Analysis, Folland Problem 1.3.15 Measures

Given a measure $\mu$ on $(X,M)$, define $\mu_0$ on $M$ by $$\mu_0(E) = \sup\{\mu(F): F\subseteq E \ \text{and} \ \mu(F) < \infty\}$$

a.) $\mu_0$ is a semifinite measure. It is called the semifinite part of $\mu$.

b.) If $\mu$ is semifinite, then $\mu = \mu_0$ (Use Exercise 14)

c.) There is a measure $\nu$ on $M$ (in general, not unique) which assumes only the values of $0$ and $\infty$ such that $\mu = \mu_0 + \nu$.

Proof:

a.) Let us prove $\mu_0$ is a semifinite measure.

i.) Since, for any $E\in M$, $\emptyset\subseteq E$ and $\mu(\emptyset)=0<+\infty$, we have that $0\in \{\mu(F): F\subseteq E \ \text{and} \ \mu(F) < \infty\}$. So $$\mu_0(E) = \sup\{\mu(F): F\subseteq E \ \text{and} \ \mu(F) < \infty\}\geqslant 0$$

ii.) Since, if $F\subseteq \emptyset$, then $F= \emptyset$. So we have
$$\mu_0(\emptyset) = \sup\{\mu(F): F\subseteq \emptyset \ \text{and} \ \mu(F) < \infty\}= \sup\{0\}=0$$

iii.) Now, let $\{E_i\}_{i\in \mathbb{N}}$ be a family of disjoint sets in $M$. Let $E=\bigcup_{i\in \mathbb{N}} E_i$. For any set $F\in M$ such that $F\subseteq E$ and $\mu(F)<+\infty$, we have that $\{F\cap E_i\}_{i\in \mathbb{N}}$ is a family of disjoint sets in $M$ and $$F=F\cap E= \bigcup_{i\in \mathbb{N}}(F\cap E_i)$$ So, $$\mu(F)=\sum_{i\in \mathbb{N}} \mu(F\cap E_i)$$ On the other hand, we have that, for all $i \in \mathbb{N}$, $F\cap E_i \subseteq E_i$ and $\mu(F\cap E_i)<+\infty$. So we have, for each $i \in \mathbb{N}$, $$ F\cap E_i \in \{H : H\subset E_i \ \text{and} \ \mu(H) < \infty\}$$ so, we have, for each $i \in \mathbb{N}$, $$\mu(F\cap E_i)\leqslant \sup \{\mu(H): H\subset E_i \ \text{and} \ \mu(H) < \infty\}=\mu_0(E_i)$$ Thus, we get $$\mu(F)=\sum_{i\in \mathbb{N}} \mu(F\cap E_i)\leqslant \sum_{i\in \mathbb{N}}\mu_0(E_i)$$ So we have $$\mu_0(E) = \sup\{\mu(F): F\subset E \ \text{and} \ \mu(F) < \infty\}\leqslant\sum_{i\in \mathbb{N}}\mu_0(E_i)$$ So we have proved $$\mu_0(E) \leqslant\sum_{i\in \mathbb{N}}\mu_0(E_i)\tag{1}$$

Now note that, if there is a $i_0 \in \mathbb{N}$ such that $\mu_0(E_{i_0})=+\infty$, then since $ E_{i_0} \subseteq E$, we have $$\{F: F\subset E_{i_0} \ \text{and} \ \mu(F) < \infty\} \subseteq \{F: F\subset E \ \text{and} \ \mu(F) < \infty\}$$ then $$+\infty = \mu_0(E_{i_0})=\sup \{\mu(F): F\subset E_{i_0} \ \text{and} \ \mu(F) < \infty\} \leqslant \\ \leqslant \sup \{\mu(F) : F\subset E \ \text{and} \ \mu(F) < \infty\}=\mu_0(E)$$ So $\mu_0(E)=+\infty$ and we have $$\mu_0(E)=+\infty =\sum_{i\in \mathbb{N}}\mu_0(E_i) \tag{2}$$

Now suppose that for all $i \in \mathbb{N}$, $\mu_0(E_i)<+\infty$ Given any $\varepsilon>0$ then, by the definition of $\mu_0$, there is $\{F_i\}_{i\in \mathbb{N}}$ a family of sets, such that $F_i \subseteq E_i$ and $$\mu_0(E_i)-\frac{\varepsilon}{2^{i+1}}\leqslant \mu(F_i)\leqslant \mu_0(E_i)<+\infty$$ Since $F_i \subseteq E_i$, we have that $\{F_i\}_{i\in \mathbb{N}}$ is a family of disjoint sets and $\bigcup_{i\in \mathbb{N}} F_i\subseteq \bigcup_{i\in \mathbb{N}} E_i=E$. So we get \begin{align} \sum_{i\in \mathbb{N}} \mu_0(E_i)-\varepsilon &=\sum_{i\in \mathbb{N}} \left(\mu_0(E_i)-\frac{\varepsilon}{2^{i+1}}\right )\leqslant \sum_{i\in \mathbb{N}} \mu(F_i)=\mu\left(\bigcup_{i\in \mathbb{N}} F_i\right) = \\ &=\sup\left\{\mu\left(\bigcup_{i=0}^k F_i\right): k\in\mathbb{N} \right\} \leqslant \\ &\leqslant\sup\{\mu(F): F\subseteq E \ \text{and} \ \mu(F) < \infty\}= \mu_0(E) \end{align} So, for any $\varepsilon>0$, $$\sum_{i\in \mathbb{N}} \mu_0(E_i)-\varepsilon\leqslant \mu_0(E)$$ So $$\sum_{i\in \mathbb{N}} \mu_0(E_i)\leqslant \mu_0(E)\tag{3}$$ From $(1)$, $(2)$ and $(3)$ we get $$\mu_0(E)=\sum_{i\in \mathbb{N}} \mu_0(E_i)$$ So $\mu_0$ is a measure.

iv.) It is easy to prove that $\mu_0$ is semifinite. In fact, given any $E\in M$ such that $$ \mu_0(E) = \sup\{\mu(F): F\subseteq E \ \text{and} \ \mu(F) < \infty\}=+\infty$$ there is at least one $ F\subseteq E$ and $\mu(F) < \infty$ such that $0<\mu(F)$, that is, $0< \mu(F) < \infty$.

b.) First, note that, for any $E\in M$, if $\mu(E)<+\infty$, then $\mu(E)=\mu_0(E)$.

Supose now that $\mu$ is semifinite. If $\mu(E)=+\infty$, we know (from Exercise 14) that for any $C\in \mathbb{R}$, $C>0$, there is $F\subseteq E$ such that $C<\mu(F)<+\infty$. So we have $$ \mu_0(E) = \sup\{\mu(F): F\subseteq E \ \text{and} \ \mu(F) < \infty\}=+\infty=\mu(E)$$

So, if $\mu$ is semifinite, then for all $E\in M$, $\mu(E)=\mu_0(E)$.

c.) Remember that, given any measure $\mu$ defined on $M$, a set $E\in M$ is $\sigma$-finite (w.r.t. $\mu$) if $E$ is the countable union of sets $E_i$ such that $\mu(E_i)<+\infty$, for all $i\in \mathbb{N}$.

Let $\nu$ be a measure on $M$ defined by, for any $E\in M$, \begin{align} \nu(E) = 0\phantom{\infty} &\;\textrm{ if $E$ is $\sigma$-finite} \\ \nu(E) = +\infty &\;\textrm{ if $E$ is not $\sigma$-finite} \end{align}

It is easy to see that $\nu$ is well defined (it is uniquely defined for all $E\in M$) and it is also easy to see that $\nu$ is a measure.

In fact, for any $E\in M$, $\nu(E)\geqslant 0$ and $\nu(\emptyset)=0$. Let $\{E_i\}_{i\in \mathbb{N}}$ be a family of disjoint sets in $M$. Let $E=\bigcup_{i\in \mathbb{N}} E_i$. If, for all $i\in \mathbb{N}$, $E_i$ is $\sigma$-finite, then $E$ is $\sigma$-finite So $$\nu(E)=0=\sum_{i\in \mathbb{N}}\nu(E_i) $$ If there is $i_0\in \mathbb{N}$, such that $E_{i_0}$ is not $\sigma$-finite, then $E$ is not $\sigma$-finite So $$\nu(E)=+\infty=\nu(E_{i_0})\leqslant \sum_{i\in \mathbb{N}}\nu(E_i) \leqslant +\infty $$ So $$\nu(E)=+\infty=\sum_{i\in \mathbb{N}}\nu(E_i) $$ So we have proved $\nu$ is a measure.

Now, it is easy to see that, for any $E\in M$, if $E$ is $\sigma$-finite, then $\mu_0(E)=\mu(E)$ and, since $\nu(E)=0$, we have
$$\mu(E)=\mu_0(E)+\nu(E)$$ On the other hand, if $E$ is not $\sigma$-finite, then $\mu(E)=+\infty$ and $\nu(E)=+\infty$. So we have $$\mu(E)=\mu_0(E)+\nu(E)$$ So we have prove $ \mu=\mu_0+\nu$.

Remark: We might be tempted to define $\nu$ by, for any $E\in M$, \begin{align} \nu(E) = 0\phantom{\infty} &\;\textrm{ if $\mu(E)<+\infty$} \\ \nu(E) = +\infty &\;\textrm{ if $\mu(E)=+\infty$} \end{align}

The issue with such definition is that this $\nu$ may not be a (countable additive) measure. In fact, suppose there is $E\in M$ such that $E$ is $\sigma$-finite and $\mu(E)=+\infty$. Then there is $\{E_i\}_{i\in \mathbb{N}}$ a family of disjoint sets in $M$ such that, for all $i \in \mathbb{N}$, $\mu(E_i)<+\infty$ and $E=\bigcup_{i\in \mathbb{N}} E_i$. We have then $\nu(E)=+\infty$ and, for all $i \in \mathbb{N}$, $\nu(E_i)=0$. So, $\nu(E)>\sum_{i \in \mathbb{N}}\nu(E_i)$.

Remark 2: The measure $\nu$ that satisfies the item c.) does not need to be unique. Here is another way to define a different $\nu$ (which we are going to call $\nu'$) that will also satisfy the item c.).

Remember that, given any measure $\mu$ defined on $M$, a set $E\in M$ is semi-finite (w.r.t. $\mu$) if, for all $F \subseteq E$, if $\mu(F)= +\infty$, there is $G\subseteq F$ such that $0<\mu(G)< \infty$. Note that if $\mu(E)<\infty$, then $E$ is trivially semi-finite.

We will also use this lemma:

Let $\{E_i\}_{i\in \mathbb{N}}$ be a family of disjoint sets in $M$. Let $E=\bigcup_{i\in \mathbb{N}} E_i$. If, for all $i\in \mathbb{N}$, $E_i$ is semi-finite, then $E$ is semi-finite.

Proof: If $F \subseteq E$ and $\mu(F)= +\infty$, then there is $i\in \mathbb{N}$, such that $\mu(F \cap E_i) >0$. We have to possibilities:

1. If $\mu(F \cap E_i) < \infty$, take $G=F \cap E_i$, then we have $G = F \cap E_i \subseteq F$ such that $0< \mu(G)< \infty$.
2. If $\mu(F \cap E_i) = \infty$, since $E_i$ is semi-finite, there is $G \subseteq F \cap E_i \subseteq F$ such that $0< \mu(G)< \infty$.

So we proved that $E$ is semi-finite. $\square$

Let $\nu'$ be a measure on $M$ defined by, for any $E\in M$, \begin{align} \nu'(E) = 0\phantom{\infty} &\;\textrm{ if $E$ is semi-finite} \\ \nu'(E) = +\infty &\;\textrm{ if $E$ is not semi-finite} \end{align}

It is easy to see that $\nu'$ is well defined (it is uniquely defined for all $E\in M$) and it is also easy to see that $\nu'$ is a measure.

In fact, for any $E\in M$, $\nu'(E)\geqslant 0$ and $\nu'(\emptyset)=0$. Let $\{E_i\}_{i\in \mathbb{N}}$ be a family of disjoint sets in $M$. Let $E=\bigcup_{i\in \mathbb{N}} E_i$. If, for all $i\in \mathbb{N}$, $E_i$ is semi-finite, then, by the lemma, $E$ is semi-finite. So $$\nu'(E)=0=\sum_{i\in \mathbb{N}}\nu'(E_i) $$ If there is $i_0\in \mathbb{N}$, such that $E_{i_0}$ is not semi-finite, then it is immediate that $E$ is not semi-finite. So $$\nu'(E)=+\infty=\nu'(E_{i_0})\leqslant \sum_{i\in \mathbb{N}}\nu'(E_i) \leqslant +\infty $$ So $$\nu'(E)=+\infty=\sum_{i\in \mathbb{N}}\nu'(E_i) $$ So we have proved $\nu'$ is a measure.

Now, it is easy to see that, for any $E\in M$, if $E$ is semi-finite, then $\mu_0(E)=\mu(E)$ and, since $\nu'(E)=0$, we have
$$\mu(E)=\mu_0(E)+\nu'(E)$$ On the other hand, if $E$ is not semi-finite, then $\mu(E)=+\infty$ and $\nu'(E)=+\infty$. So we have $$\mu(E)=\mu_0(E)+\nu'(E)$$ So we have prove $ \mu=\mu_0+\nu'$.

It is easy to see that $\nu'$ is, in general, different from the measure $\nu$ that we defined in c.). In fact, if $E \in M$ is semi-finite but not $\sigma$-finite, we have $\nu'(E)=0 < +\infty = \nu(E)$.