Exercise 1.2.23 (Uniqueness of Lebesgue measure)

Question

Let $\mathcal{L}\left(\mathbb{R}^dight)$ be the set of all Lebesgue measurable sets on $\mathbb{R}^d$, and let
$$m':\mathcal{L}\left(\mathbb{R}^dight)	o[0,+\infty], \quad E \mapsto m'(E)$$
be a map that obeys the following axioms:
\begin{enumerate}[label=(oman*)]
	\item (Empty set) $m(\emptyset)=0$.
	\item (Countable additivity) If $E_1,E_2,\ldots \subseteq \mathbb{R}^d$ is a contable sequence of disjoint Lebesgue measurable sets, then $m(\bigcup_{n=1}^{\infty}E_n) = \sum_{n=1}^{\infty}m(E_n)$.
	\item (Translation invariance) If $E$ is Lebesgue measurable and $x\in\mathbb{R}^d$ then $m(E+x) = m(E)$.
	\item (Normalisation) $m([0,1]^d) = 1$
\end{enumerate}
Show that $m'=m$.

Elu Thingol · Accepted Answer

We first add extra properties to the list that we already have
\begin{enumerate}[label=(oman*)]
	\item[(v)] (Monotonicity) If $E_1 \subseteq E_2 \subseteq \ldots \subseteq \mathbb{R}^d$ is a countable sequence of Lebesgue measurable sets, then $m(E_1) \leq m(E_2) \leq \ldots$
	\item[(vi)] (Countable subadditivity) If $E_1,E_2,\ldots \subseteq \mathbb{R}^d$ is a contable sequence of Lebesgue measurable sets, then $m(\bigcup_{n=1}^{\infty}E_n) \leq \sum_{n=1}^{\infty}m(E_n)$.
\end{enumerate}
Monotonicity follows from additivity, since $E_n = E_{n-1}\cup (E_n/E_{n-1})$ and thus $m(E_n) = m(E_{n-1}) + m(E_n/E_{n-1}) \geq m(E_{n-1})$. The subadditivity can be shown using additivity for finite unions $m(\bigcup_{n=1}^{N}E_n) \leq \sum_{n=1}^{N}m(E_n)$. Using the convergence therems allows us to take $N	o\infty$ and obtain the coutable version.
ewline

Now we have to demonstrate that $\forall E \in \mathcal{L}\left(\mathbb{R}^dight): m(E) = m'(E)$. We use the following strategy: we first demonstrate it for elementary sets, then for open sets, and only then do we generalise to the Lebesgue measurable sets.
  \begin{itemize}
	\item Elementary sets
ewline
		Note that $m(E) = m'(E)$ for all $A\in\mathcal{E}(\mathbb{R}^d)$ follow directly from \href{./exercise_1-1-3}{Exercise 1.1.3}, since both $m$ and $m'$ satisfy the axioms of elementary measure together with the normalisation property.
	\item Bounded open sets
ewline
		Let $U$ be an open set with $m(U) < \infty$. By famous Lemma 1.2.11 we write $U$ as a countable union of (almost) disjoint boxes $\bigcup_{n=1}^{\infty}B_n$. We can make them disjoint by setting the boundary $F := \bigcup_{n=1}^{\infty}B_n / \bigcup_{n=1}^{\infty}B^\circ_n$. By additivity of the both functions we want to obtain the following equality:
    $$m(U) = m(F) + m\left(\bigcup_{n=1}^{\infty}B^\circ_night) \stackrel{??}{=}  m'(F) + m'\left(\bigcup_{n=1}^{\infty}B^\circ_night) = m'(U)$$ 
    First notice $F$ is a Lebesgue null set $m(F) = m(U) - m(\bigcup_{n=1}^{\infty}B^\circ_n) = m(\bigcup_{n=1}^{\infty}B_n) - m(\bigcup_{n=1}^{\infty}B^\circ_n) = \sum_{n=1}^{\infty}B_n - \sum_{n=1}^{\infty}B^\circ_n = 0$. But $m$ and $m'$ always agree on Lebesgue null sets. If $F$ is a null set, then we can find a box cover $\bigcup_{n=1}^{\infty}C_n$ such that $\sum_{n=1}^{\infty}|C_n| \leq \epsilon$. Since $m$ and $m'$ agree on elementary sets, we have by subadditivity $m'(E) \leq m'(\bigcup_{n=1}^{\infty}C_n) \leq \sum_{n=1}^{\infty}m'(C_n) = \sum_{n=1}^{\infty}|C_n| \leq \epsilon$. Thus, $m(F) = m'(F) = 0$. The latter terms must also agree, since this is a disjoint union of elementary sets, and the equality holds indeed./li>
			\item Bounded Lebesgue measurable sets
ewline
			Let $E$ be a bounded Lebesgue measurable set with $m(E) < \infty$. We demonstrate that $m(E) = m'(E)$. Since both maps fulfill very similar properties, we give ourselves an epsilon of a room.
			
			\begin{itemize}\item[$\geq$)] We use the outer regularity of the Lebesgue measure. Since $E$ is bounded, we can a bounded open set $U$ containing $E$ such that $m(U) - m(E) = m(U/E) \leq \epsilon$. We then have $m'(E) \leq m'(U) = m(U) \leq m(E) + \epsilon$. Since our choice of $\epsilon$ was arbitrary, $m'(E) \leq m(E)$.
	\item[$\leq$)] Since $E$ is bounded, we can contain it in a box $B \supseteq E$. By additivity of $m'$ we have $m'(B) = m'(E) + m'(B/E)$ and thus
		$$m(E) = m(B) - m(B/E) = m'(B) - m(B/E) = m'(E) + m'(B/E) - m(B/E) \leq m'(E)$$
		where in the last step we used the same trick as in the first part of this proof to demonstrate that $m(B/E) \geq m'(B/E)$.
\end{itemize}
\end{itemize}

To generalise this to any unbounded Lebesgue measurable set $E$, write $E_n = E \cap B(0,n) / E_{n-1}$ where $B(0,n)$ is a ball of radius $n$ around the origin. We then have by the countable additivity of both measures
$$m(E) = m\left(\bigcup_{n=1}^{\infty}E_night) = \sum_{n=1}^{\infty}m(E_n) = \sum_{n=1}^{\infty} m'(E_n) = m'\left(\bigcup_{n=1}^{\infty}E_night) = m'(E).$$

Prakash Anand · Answer

\begin{proof} We already know that $m^{\prime}(E)=m(E)$ for any Jordan measurable set $E$, and specifically for boxes.\par It is simple to prove that $m^{\prime}$ is sub-additive and monotone.\par extbf{Step I.} We show that for $A \in \mathcal{L}$, if $m(A)=0$ then $m^{\prime}(A)=0$ : Let $A$ be a Lebesgue measurable set of 0 measure, $m(A)=0$. For any $\varepsilon>0$, let $\left(B_{n} ight)_{n}$ be a sequence of boxes such that $A \subset \bigcup_{n} B_{n}$ and such that $\sum_{n}\left|B_{n} ight| \leq \varepsilon$. Then, $m^{\prime}(A) \leq \sum_{n} m^{\prime}\left(B_{n} ight)=$ $\sum_{n}\left|B_{n} ight| \leq \varepsilon$, and taking $\varepsilon ightarrow 0$, we have that $m^{\prime}(A)=0$.\par extbf{Step II}. We show that if $U$ is open of finite Lebesgue measure, then $m^{\prime}(U)=m(U)$ : Let $U$ be an open set such that $m(U)<\infty$. Write $U=\bigcup_{n} B_{n}$ where $\left(B_{n} ight)_{n}$ are almost disjoint closed boxes. Let $B_{n}^{\prime}=\left(B_{n} ight)^{\circ}$, so $\left(B_{n}^{\prime} ight)_{n}$ are pairwise disjoint. Let $F=\bigcup_{n} B_{n} \backslash \biguplus_{n} B_{n}^{\prime}$. Since $$ m(F)=m(U)-m\left(\biguplus_{n} B_{n}^{\prime} ight)=\sum_{n}\left|B_{n} ight|-\sum_{n}\left|B_{n}^{\prime} ight|=0, $$ we have that $m^{\prime}(F)=0$. Note that $U=F \uplus \biguplus_{n} B_{n}^{\prime}$, so $$ m^{\prime}(U)=m^{\prime}(F)+\sum_{n} m^{\prime}\left(B_{n}^{\prime} ight)=\sum_{n}\left|B_{n}^{\prime} ight|=m(U) . $$ extbf{Step III}. We show that for any $A \in \mathcal{L}$, the inequality $m^{\prime}(A) \leq m(A)$ holds: Let $A$ be a Lebesgue measurable set with $m(A)<\infty$. Fix $\varepsilon>0$ and let $U$ be an open set such that $A \subset U$ and $m(U)-m(A)=m(U \backslash A)<\varepsilon$. Then, $m^{\prime}(A) \leq m^{\prime}(U) \leq m(A)+\varepsilon$. Taking $\varepsilon ightarrow 0$ we have that for any $A \in \mathcal{L}$, the inequality $m^{\prime}(A) \leq m(A)$ holds. (The case where $m(A)=\infty$ is immediate.) extbf{Step IV}. We show that for any bounded set $A \in \mathcal{L}$ we have $m^{\prime}(A)=m(A)$ : Let $A \in \mathcal{L}$ be a bounded set. Let $B$ be a box bounding $A \subset B$. We have $m^{\prime}(B \backslash A)=m^{\prime}(B)-m^{\prime}(A)$ by additivity of $m^{\prime}$. Also, $$ m^{\prime}(A) \leq m(A)=m(B)-m(B \backslash A) \leq m^{\prime}(B)-m^{\prime}(B \backslash A)=m^{\prime}(A) . $$ extbf{Step V}. We show that for any $A \in \mathcal{L}$ we have $m^{\prime}(A)=m(A)$ : Let $A \in \mathcal{L}$. Write $A=\biguplus A_{n}$ where $A_{n}=B_{n} \backslash B_{n-1}$, and $B_{n}=[-n, n]^{d} .$ Then by additivity of both $m, m^{\prime}$ and since $A_{n}$ are all bounded, $$ m^{\prime}(A)=\sum_{n} m^{\prime}\left(A_{n} ight)=\sum_{n} m\left(A_{n} ight)=m(A). $$ \end{proof}

Exercise 1.2.23 (Uniqueness of Lebesgue measure)

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Exercise 1.2.23 (Uniqueness of Lebesgue measure)

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