2. Construction of Lebesgue Measure

$\newcommand{\argmin}{\mathop{\mathrm{argmin}}\limits}$ $\newcommand{\argmax}{\mathop{\mathrm{argmax}}\limits}$

In this chapter, we construct the Lebesgue measure on $\mathbb{R}^d.$ For this, we prove Riesz representation theorem and use the result to construct a complete measure space $(\mathbb{R}, \mathfrak{M}, m)$ such that integration with respect to $m$ is equal to Riemann integration for all Riemann-integrable functions. We then use $\sigma$-compactness of $\mathbb{R}$ to show that such $m$ is the Lebesgure measure.

Throughout chapter 2, without any further description, I will denote compact sets by $K$ or $H,$ open sets by $U$ or $V$ and closed sets by $F.$

Topological preliminaries

Hausdorff space

A space $X$ is Hausdorff, if for all $x,y \in X,$ there exist $U,V \sub X$ such that $x \in U,$ $y \in V$ and $U \cap V = \phi.$

For a Hausdorff $X$ and $\{K_i\}_{i \in I}$ such that $\cap_{i\in I} K_i = \phi,$ $$ \exists \text{ a finite }J \text{ s.t. } \cap_{i\in J} K_i = \phi. $$

Locally compactness

$X$ is locally compact, if $$ \forall x \in X, \exists U \text{ such that } x\in U \text{ and } \overline U \text{ is compact.} $$

Locally compact & Hausdorff

(i) $$ C_c(X) := \{f:X\to\mathbb{R},~ f \text{ is continuous and } \text{supp.}f \text{ is compact}\} $$ where $\text{supp.}f:= \overline{\{ f > 0\}}.$
(ii) We write $K \prec f,$ if $f \in C_c(X),$ $0 \le f \le 1$ and $f=1$ on $K.$
(iii) We write $f \prec V,$ if $f \in C_c(X),$ $0 \le f \le 1$ and $\text{supp.}f \sub V.$

The following three theorems are being used frequently.

For a locally compact Hausdorff space $X$ and $K \sub U \sub X,$ $$ \exists V \text{ such that } K \sub V \sub \overline V \sub U. $$

For a locally compact Hausdorff space $X$ and $K \sub V \sub X,$ $$ \exists f \in C_c(X) \text{ such that } K \prec f \prec V. $$

For a locally compact Hausdorff space $X,$ $K \sub X$ and open cover $\{V_1,\cdots,V_n\}$ of $K,$ $$ \exists h_1,\cdots,h_n \in C_c(X) \text{ such that } h_i \prec V_i, ~\forall i \text{ and } K \prec \sum_{i=1}^n h_i. $$

Riesz representation theorem

Riesz representation theorem (RRT) states that positive linear functional defined on $C_c$ of locally compact Hausdorff space can be represented as Lebesgue integral with respec to some complete measure space.

Let $X$ be a locally compact Hausdorff space, and $\Lambda: C_c(X) \to \mathbb{R}$ be a positive linear functional. Then there exists a $\sigma$-algebra $\mathfrak{M}$ that contains all Borel sets and uniquely exists a measure $\mu$ that satisfies the followings.
(i) $\Lambda f = \int_X f d\mu,$ for all $f \in C_c(X).$
(ii) $\mu(K) < \infty,$ for all compact $K.$
(iii) $\mu(E) = \inf\{\mu(V): E \sub V,~ V \text{ is open}\},$ for all $E \in \mathfrak{M}.$
(iv) $\mu(E) = \sup\{\mu(K): K\sub E,~ K \text{ is compact }\},$ for all $E \in \mathfrak{M}$ such that $\mu(E) < \infty$ for $E$ is open.
(v) $(X, \mathfrak{M}, \mu)$ is complete.
(i) The representation is well-defined since $f$ is Borel measurable beause it is continuous.
(ii) Note that the codomain of $\Lambda$ is $\mathbb{R},$ which implies $\Lambda f < \infty$ for all $f \in C_c(X).$
(iii) This is called outer regularity, which means measure of some sets can be approximated by larger open sets.
(iv) This is called inner regularity, which means measure of some sets can be approximated by smaller compact sets.
(+) We say $(X, \mathfrak{M}, \mu)$ is regular, if all Borel sets are inner and other regular. We will later show that if $X$ is in addition a $\sigma$-compact space, then $(X, \mathfrak{M}, \mu)$ is regular.

Proof of RRT

Proof of the theorem is quite long and exhaustive. I would like to divide it into 11 steps.

The first step is to prove uniqueness of such $\mu.$ The last step is to prove that such measure space satisfies $\Lambda f = \int_X f d\mu.$ These two steps are not as intercorrelated as the steps inbetween, so I will only number the 9 steps inbetween these two.

We define

\[\mu(V) := \sup\{\Lambda f: f \prec V\}, \\ \mu(E) := \inf\{\mu(V): E \sub V\}, \\ \mathfrak{M}_F := \left\{E \sub X: \mu(E) > \infty,~ \mu(E) = \sup\{\mu(K): \text{compact }K \sub E\}\right\} \\ \mathfrak{M} := \{E \sub X: E \cap K \in \mathfrak{M}_F,~ \forall \text{ compact } K\}.\]

Then show the followings one at a time.

  1. $\mu\left( \cup_{i=1}^\infty E_i \right) \le \sum_{i=1}^\infty \mu(E_i)$ for all $E_i \sub X.$
  2. $\mu(K) = \inf \{\Lambda f: K \prec f\}.$
    Hence $\mu(K) < \infty$ and $K \in \mathfrak{M}_F.$
  3. Any open $V$ is inner regular.
    Hence if $\mu(V) < \infty,$ then $V \in \mathfrak{M}_F.$
  4. $\mu\left( \cup_{i=1}^\infty E_i \right) = \sum_{i=1}^\infty \mu(E_i)$ for disjoint $E_i \in \mathfrak{M}F.$ And if $\mu(\cup{i=1}^\infty E_i) < \infty,$ then $\cup_{i=1}^\infty E_i \in \mathfrak{M}_F.$
  5. For a given $E \in \mathfrak{M}_F$ and $\epsilon > 0,$ there exists $K, V$ such that $K \sub E \sub V$ and $\mu(V \setminus K) < \epsilon.$
  6. If $A, B \in \mathfrak{M}_F,$ then $A\setminus B,$ $A \cap B,$ $A\cup B \in \mathfrak{M}_F.$
  7. $\mathfrak{M}$ is a $\sigma$-algebra containing all Borel sets.
  8. $\mathfrak{M}_F = \{E\in\mathfrak{M}: \mu(E)< \infty\}.$
  9. $\mu\left( \cup_{i=1}^\infty E_i \right) = \sum_{i=1}^\infty \mu(E_i)$ for disjoint $E_i \in \mathfrak{M}.$

Uniqueness of $\mu$

Suppose $\mu_1, \mu_2$ are measures that satisfies the conditions.

Claim 1: If $\mu_1(K)=\mu_2(K)$ for all compact $K,$ then $\mu_1(E) = \mu_2(E)$ for all $E \in \mathfrak{M}.$

Given an open $V,$ $$ \begin{aligned} \mu_1(V) &= \sup \left\{ \mu_1(K): \text{compact } K \sub V \right\} \\ &= \sup \left\{ \mu_2(K): \text{compact } K \sub V \right\} \\ &= \mu_2(V). \end{aligned} $$ Now, given an $E \in \mathfrak{M},$ $$ \begin{aligned} \mu_1(E) &= \sup \left\{ \mu_1(V): \text{open } V \supset E \right\} \\ &= \sup \left\{ \mu_2(V): \text{open } V \supset E \right\} \\ &= \mu_2(E). \end{aligned} $$

Claim 2: $\mu_1(K) = \mu_2(K),~ \forall \text{ compact }K.$

For a given $\epsilon > 0,$ by outer regularity there exists $V$ such that $K \sub V$ and $\mu_2(V) \le \mu_2(K) + \epsilon.$ By Urysohn's lemma, there exists $f$ such that $K \prec f \prec V.$ $$ \begin{aligned} \mu_1(K) &= \int_X \chi_K d\mu_1 \\ &\le \int_X f d\mu_1 = \Lambda f \\ &= \int_X f d\mu_2 \\ &\le \int_X \chi_V d\mu_2 \\ &= \mu_2(V) \le \mu_2(K) + \epsilon. \end{aligned} $$ Similarly, one can show that $\mu_2(K) \le \mu_1(K) + \epsilon.$ By letting $\epsilon \downarrow 0,$ we get the result.

1. $\mu\left( \cup_{i=1}^\infty E_i \right) \le \sum_{i=1}^\infty \mu(E_i)$ for all $E_i \sub X$

Partition of unity plays an important part of the proof.

(1) $\mu(V_1 \cup V_2) \le \mu(V_1) + \mu(V_2).$

Given $g \prec V_1 \cup V_2,$ and let $K=\text{supp.}g$ be a compact set. Then by partition of unity, there exists $h_1,h_2 \in C_c(X)$ such that $h_i \prec V_i,$ $i=1,2$ and $K \prec h_1 + h_2.$ Thus $$ g(x) = g(x)\left(h_1(x) + h_2(x)\right), \\ \Lambda g = \Lambda(g\cdot h_1) + \Lambda(g \cdot h_2) $$ Take supremum to $g$ on both sides, and we get $$ \mu(V_1 \cup V_2) \le \mu(V_1) + \mu(V_2). $$

(2) By induction, $\mu(\cup_{i=1}^n V_i) \le \sum_{i=1}^n \mu(V_i).$

(3) $\mu\left( \cup_{i=1}^\infty E_i \right) \le \sum_{i=1}^\infty \mu(E_i).$

For given $\epsilon >0,$ there exists $V_i \supset E_i$ such that $\mu(V_i) \le \mu(E_i) + \frac{\epsilon}{2^i}.$ Take $f$ such that $f \prec \cup_{i=1}^\infty V_i.$ Since $\text{supp.}f$ is compact, there exists $n$ such that $f \prec \cup_{i=1}^n V_i.$ $$ \begin{aligned} \Lambda f &= \mu\left( \cup_{i=1}^n V_i \right) \\ &\le \sum_{i=1}^n \mu(V_i) \\ &\le \sum_{i=1}^\infty \mu(V_i) \\ &\le \sum_{i=1}^\infty \mu(E_i) + \epsilon. \end{aligned} $$ Taking supremum to $f$ to both sides gives $$ \mu\left(\cup_{i=1}^\infty E_i\right) \le \mu\left(\cup_{i=1}^\infty V_i\right) \le \sum_{i=1}^\infty \mu(E_i) + \epsilon. $$ Letting $\epsilon \downarrow 0$ gives the desired result.

2. $\mu(K) = \inf \{\Lambda f: K \prec f\}$

It is equivalent to show that (1) $\mu(K) \le \Lambda f$ for all $f: K \prec f$ and (2) for given $\epsilon > 0,$ there exists $f \prec K$ such that $\mu(K) \ge \Lambda f - \epsilon.$

(1) $\mu(K)$ is a lower bound of $\Lambda f,$ $K \prec f.$

Given $f \succ K,$ let $V_\alpha = \{f > \alpha\}$ for $\alpha < 1.$ Then since $f$ is continuous, $V_\alpha$ is open. Since $f > \alpha$ on $V_\alpha,$ $\alpha g \le f$ hence $\alpha \Lambda f \le \Lambda f$ for $g \prec V_\alpha.$ $$ \mu(K) \le \mu(V_\alpha) = \sup \{\Lambda g: g \prec V_\alpha\} \le \frac{1}\alpha \Lambda f. $$ Letting $\alpha \to 1,$ we get the result.

(2) $\mu(K)$ is the greatest lower bound of $\Lambda f$, $K \prec f.$

Given $\epsilon > 0$. By outer regularity, there exists $V$ such that $K \sub V$ and $\mu(V) \le \mu(K) + \epsilon.$ By Urysohn's lemma, there exists $f$ such that $K \prec f \prec V.$ By definition of $\mu,$ $$ \begin{aligned} \Lambda f &\le \mu(V) \\ &\le \mu(K) + \epsilon. \end{aligned} $$

Hence $\mu(K) \le \Lambda f \in \mathbb{R}$ thus is finite and $K \in \mathfrak{M}_F.$

3. $V$ is inner regular

Given $\epsilon > 0.$ It suffices to show that there exists $K \sub V$ such that $\mu(V) - \epsilon \le \mu(K).$ Let $\alpha = \mu(V) - \epsilon,$ then by definition of $\mu,$ there exists $f$ such that $f \prec V$ and $\Lambda f > \alpha.$ Let $K = \text{supp.} f$ be a compact set. To show that $\mu(K) \ge \Lambda f,$ we will use the fact $$ \mu(K) = \inf\{\mu(U): K \sub U\} \text{ and} \\ \mu(U) = \sup\{\Lambda g: g \prec U\}. $$ Take $U \supset K,$ then by definition $f \prec U$ and $$ \Lambda f \le \mu(U) := \sup\{\Lambda g: g \prec U\}. $$ Thus by taking infimum to $\mu(U),$ $U \supset K$ to both sides, we get $$ \Lambda f \le \inf \{\mu(U): U \supset K\} = \mu(K). $$

Hence if $\mu(V) < \infty,$ then $V \in \mathfrak{M}_F.$

4. $\mu$ is $\sigma$-additive on $\mathfrak{M}_F~$ & $~\mu(E) < \infty \Rightarrow E \in \mathfrak{M}_F$

Since we already proved 1., it suffices to show the other side of inequality. For this, we will divide the proof into substeps similar to that in 1.

(1) $\mu(K_1 \cup K_2) \ge \mu(K_1) + \mu(K_2),$ $K_1,K_2$ are disjoint.

Observe that $K_1 \sub K_2^c$ and $K_2^c$ is open. By Urysohn's lemma, there exists $f$ such that $K_1 \prec f \prec K_2^c.$ Using the result from 2., for given $\epsilon >0$ there exists $g$ such that $K_1 \cup K_2 \prec g$ and $\Lambda g \le \mu(K_1 \cup K_2) + \epsilon.$ Thus $$ K_1 \prec f\cdot g \text{ and } K_2 \prec (1-f)\cdot g $$ and $$ \begin{aligned} \mu(K_1) + \mu(K_2) &\le \Lambda(f\cdot g) + \Lambda((1-f)\cdot g) \\ &= \Lambda g \le \mu(K_1 \cup K_2) + \epsilon. \end{aligned} $$ By letting $\epsilon \to 0,$ the result follows.

(2) By induction, $\mu\left(\cup_{i=1}^n K_i\right) \ge \sum_{i=1}^n \mu(K_i)$ for disjoint $K_i$’s

(3) $\mu\left( \cup_{i=1}^\infty E_i \right) \ge \sum_{i=1}^\infty \mu(E_i),$ $E_i \in \mathfrak{M}_F$ are disjoint.

For $E_i$, by inner regularity there exists $K_i \sub E_i$ such that $\mu(K_i) \ge \mu(E_i) - \frac{\epsilon}{2^i}.$ $$ \begin{aligned} \mu\left( \cup_{i=1}^\infty E_i \right) &\ge \mu\left( \cup_{i=1}^n E_i \right) \\ &\ge \mu\left( \cup_{i=1}^n K_i \right) \\ &\ge \sum_{i=1}^n \mu(K_i) \\ &\ge \sum_{i=1}^n \mu(E_i) - \epsilon. \end{aligned} $$ By letting $\epsilon \to 0,$ we get the result.

If $\mu(E) < \infty,$ then for all $\epsilon>0,$ there exists $N\in\mathbb{N}$ such that

\[\begin{aligned} \mu(E) &\le \sum_{i=1}^N \mu(E_i) + \epsilon \\ &\le \sum_{i=1}^N \mu(K_i) +2\epsilon \\ &= \mu\left(\cup_{i=1}^N K_i\right) + 2\epsilon. \end{aligned}\]

This implies $E$ is inner regular, hence $E \in \mathfrak{M}_F.$

5. $\forall E \in \mathfrak{M}_F,$ $\forall \epsilon > 0,$ $\exists K, V$ such that $K \sub E \sub V$ and $\mu(V \setminus K) < \epsilon$

By outer regularity, $$ \exists V \supset E \text{ such that } \mu(V) \le \mu(E) + \epsilon/2. $$ By inner regularity, $$ \exists K \sub E \text{ such that } \mu(E) \le \mu(K) + \epsilon/2. $$ Since $K \sub V,$ note that $V = K \cup (V\setminus K).$ $$ \begin{aligned} K \text{ is compact} &\implies K \in \mathfrak{M}_F,\\ V\setminus K \text{ is open and }&\\ \mu(V\setminus K) \le \mu(V) < \mu(K) + \epsilon < \infty &\implies V \setminus K \in \mathfrak{M}_F. \end{aligned} $$ By 4., it suffices to show that $\mu(V) - \mu(K) < \epsilon.$ $$ \mu(V) = \mu(K) + \mu(V \setminus K), \\ \mu(V \setminus K) = \mu(V) - \mu(K) \le \epsilon. $$

6. If $A, B \in \mathfrak{M}_F,$ then $A\setminus B,$ $A \cap B,$ $A\cup B \in \mathfrak{M}_F$

(1) $A \setminus B \in \mathfrak{M}_F.$

Given $\epsilon > 0,$ by 5. there exist $K_1 \sub A \sub V_1$ and $K_2 \sub B \sub V_1$ such that $\mu(V_1\setminus K_1) < \epsilon $ and $\mu(V_2 \setminus K_2) < \epsilon.$ Then $$ A \setminus B \sub (V_1 \setminus K_1) \cup (V_2 \setminus K_2) \cup (K_1 \setminus V_2). $$ By 1., $$ \begin{aligned} \mu(A \setminus B) &\le \mu(V_1 \setminus K_1) + \mu(V_2 \setminus K_2) + \mu(K_1 \setminus V_2) \\ &\le \mu(K_1 \setminus V_2) + 2\epsilon < \infty \end{aligned} $$ since $K := K_1 \setminus V_2$ is compact. This implies $A\setminus B$ is inner regular and its measure is finite, thus $A \setminus B \in \mathfrak{M}_F.$

(2) $A\cap B, A\cup B \in \mathfrak{M}_F$

$$ A \cap B = A \setminus(A\setminus B) \in \mathfrak{M}_F, \\ A \cup B = A \cup (B\setminus A) \in \mathfrak{M}_F. $$

7. $\mathfrak{M}$ is a $\sigma$-algebra containing all Borel sets

It is not difficult to show that $\mathfrak{M}$ is a $\sigma$-algebra. Now it is enough to show that all closed sets are contained in $\mathfrak{ M}.$ Given a closed $F,$ $$ \begin{aligned} F \cap K \subset K &\implies F \cap K \text{ is compact}, \forall K \\ &\implies F \cap K \in \mathfrak{M}_F, \forall K \end{aligned} $$

8. $\mathfrak{M}_F = \{E\in\mathfrak{M}: \mu(E)< \infty\}$

It is clear that $E \in \mathfrak{M}_F$ implies $E \in \mathfrak{M}.$

ETS: $E \in \mathfrak{M},$ $\mu(E) < \infty.$ $\implies$ $E$ is inner regular.

Given $\epsilon > 0,$ there exists $V \supset E$ such that $\mu(V) \le \mu(E) + \epsilon < \infty.$ Thus such $V \in \mathfrak{M}_F.$ By 5., there exists $K \subset V$ such that $\mu(V \setminus K) \le \epsilon.$
For such $K,$ by definition of $\mathfrak{M},$ $E \cap K \in \mathfrak{M}_F.$ Then by inner regularity, there exists a compact $H \sub (E\cap K)$ such that $$ \mu(E\cap K) \le \mu(H) + \epsilon. $$ Note that $E \sub (E \cap K) \cup (V \setminus K),$ thus $$ \mu(E) \le \mu(E\cap K) + \mu(V\setminus K) \le \mu(K) + 2\epsilon $$ which makes $E \in \mathfrak{M}_F.$

9. $\mu$ is a measure on $(X, \mathfrak{M})$

ETS: $\mu\left( \cup_{i=1}^\infty E_i \right) = \sum_{i=1}^\infty \mu(E_i)$ for disjoint $E_i \in \mathfrak{M}.$

If $\mu(E_i) = \infty$ for some $E_i,$ then it is trivial.
Otherwise, it implies $E_i \in \mathfrak{M}_F$ for all $i=1,2,\cdots.$ Thus by 4., it is clear.

$\Lambda f = \int_X f d\mu$ for all $f \in C_c(X)$

Observe that if $\Lambda f \le \int_X fd\mu,$ then by applying this to $-f$ we get $\Lambda f \ge \int_X f d\mu.$ Thus it suffices to show that $\Lambda f \le \int_X f d\mu.$

Given $f \in C_c(X),$ let $K=\text{supp.}f$ be a compact set. Since $f$ is continuous, by min-max theorem $f(X) = f(K)$ is compact in $\mathbb{R}$. Thus there exists $a, b \in \mathbb{R}$ such that $f(K) \sub (a, b].$

For a small enough $\epsilon > 0,$ Let $$ a=y_0<y_1<\cdots<y_n = b,\\ y_i-y_{i-1} < \epsilon,~ \forall 1\le i\le n. $$

Let $E_i = f^{-1}(y_{i-1}, y_i] \cap K \in \mathfrak{M}$ so that $\{E_1,\cdots,E_n\}$ be a partition of $K.$ Then by Outer regularity, there exist open $W_i$'s such that $$ E_i \sub W_i,~ \mu(W_i) \le \mu(E_i) + \epsilon/n. $$ In addition, let $$ U_i = f^{-1}(y_{i-1}, y_i+\epsilon),~ 1\le i\le n $$ be open sets. Finally, let $$ V_i = W_i \cap U_i. $$ Then clearly the followings hold. $$ \begin{aligned} &\text{(1) } E_i \sub V_i,~ \forall i \\ &\text{(2) } f \in (y_{i-1}, y_i) \text{ on } V_i,~ \forall i \\ &\text{(3) } \mu(V_i) \le \mu(E_i) + \epsilon / n \end{aligned} $$

(1) implies that $(V_1,\cdots, V_n)$ is an open cover of $K.$ By partition of unity, there exist $h_1,\cdots,h_n \in C_c(X)$ such that $h_i \prec V_i$ for all $i$ and $K \prec \sum_{i=1}^n h_i.$ With the result from 2., this implies $$ \mu(K) \le \Lambda \left( \sum_{i=1}^n h_i \right) $$ Note that since $K$ is the support of $f,$ $$ f = f\cdot \sum_{i=1}^n h_i = \sum_{i=1}^n f\cdot h_i $$ and since $h_i \prec V_i,$ $$ f \cdot h_i \le (y_i + \epsilon)\cdot h_i. $$ Thus $$ \begin{aligned} \Lambda f =& \sum_{i=1}^n \Lambda \left( f \cdot h_i \right) \\ \le& \sum_{i=1}^n (y_i + \epsilon)\Lambda (h_i) \\ =& \sum_{i=1}^n (y_i + \epsilon + |a|) \Lambda (h_i) - |a|\sum_{i=1}^n \Lambda(h_i) \\ \le& \sum_{i=1}^n (y_i + \epsilon + |a|) \mu (V_i) - |a| \Lambda\left(\sum_{i=1}^n h_i\right) \\ \le& \sum_{i=1}^n (y_i + \epsilon + |a|)\left(\mu(E_i) + \frac{\epsilon}{n}\right) - |a|\mu(K) \\ =& \sum_{i=1}^n (y_i-\epsilon)\cdot\mu(E_i) + 2\epsilon\mu(K) + \cancel{|a|\mu(K)} \\ &+ \frac{\epsilon}{n}\sum_{i=1}^n (y_i + \epsilon + |a|) \\ &- \cancel{|a|\mu(K)} \\ \le& \int_X f d\mu + \epsilon\left( 2\mu(K) + |b| + |a| + \epsilon \right). \end{aligned} $$ Letting $\epsilon\to0$ yields the desired result.

(contents below are crudely written: the rationale, proofs, implication of theorems will be added later)

Regularity of measures


Let $(X, \mathfrak{M}, \mu)$ be the measure space constructed from RRT. If $E \in \mathfrak{M}$ is $\sigma$-finite, then $E$ is inner regular.


Let $(X, \mathfrak{M}, \mu)$ be the measure space constructed from RRT. If $X$ is $\sigma$-compact, then the following hold.
(i) For $E \in \mathfrak{M}$ and $\epsilon>0,$ there exists a closed $F$ and and open $V$ such that $F \sub E \sub V$ and $\mu(V-F)<\epsilon.$
(ii) All $E \in \mathcal{B}$ is $\sigma$-finite, hence $\mu$ is regular.
(iii) For $E \in \mathfrak{M},$ there exists a $F_\sigma$-set $A$ and and $G_\delta$-set $B$ such that $A \sub E \sub B$ and $\mu(B-A)=0.$

Let $(X, \mathcal{B}, \lambda)$ be a measure space defined on a Borel $\sigma$-algebra $\mathcal{B}.$ If all open sets $V$ are $\sigma$-compact and $\lambda(K) < \infty$ for all compact $K,$ then $\lambda$ is regular.

The Lebesgue measure


$W:= \prod_{i=1}^d \langle a_i, b_i \rangle,$ where $\langle\cdot,\cdot\rangle$ can be either one of open, closed or half-open interval, is a cell in $\mathbb{R}^d.$

There exists a measure space $(\mathbb{R}^d, \mathfrak{M}, m)$ that satisfies the followings.
(i) $(\mathbb{R}^d, \mathfrak{M}, m)$ is a complete regular measure space, and $$ \forall E \in \mathfrak{M},~ \exists A \in F_\sigma, B \in G_\delta \text{ s.t. } A \sub E \sub B,~ m(B-A)=0. $$ (ii) $m(W) = \text{vol}(W)$ for all cells $W.$

(i) RRT & $\sigma$-compactness of $\mathbb{R}^d.$
(ii) (1) open cells ($\sigma$-compact -> define $g_n \uparrow \chi_V$. + convergence in two perspective). (2) all cells

(continued from the theorem above)
(iii) $m(E) = m(E+x)$ for all $x \in X.$
(iv) If $\mu$ is a measure on $(\mathbb{R}^d, \mathfrak{M})$ that is translation invariant and $\mu(K)<\infty$ for all compact $K,$ then $\mu(E) = c\cdot m(E)$ for all $E\in\mathcal{B}$ for some $c\in\mathbb{R}.$
(v) For a linear map $T: \mathbb{R}^d \to \mathbb{R}^d,$ there exists $\Delta_T \in \mathbb{R}$ such that $m(T(E)) = \Delta_T\cdot m(E)$ for all $E \in \mathcal{B}.$


  • Rudin. 1986. Real and Complex Analysis. 3rd edition. McGraw-Hill.
  • Real Analysis (Spring, 2021) @ Seoul National University, Republic of Korea (instructor: Prof. Insuk Seo).