> [!definition] > > Let $F(x) = x$, then the [[Lebesgue-Stieltjes Measure]] $m = \mu_F$ associated with $F$ is the **Lebesgue measure**. > > The collection $\cm_\mu$ of Lebesgue-measurable sets is denoted as $\cl$. > [!theorem] > > Let $E \subset \real$, and let > $ > s + E = \bracs{s + x: x \in E} \quad > rE = \bracs{rx: x \in E} > $ > If $E \in \cl$, then $r + E, rE \in \cl$ are also Lebesgue-measurable. Moreover, $m(E) = m(r + E)$ and $m(rE) = |r|m(E)$. > > In other words, $m$ is invariant under translations and dilations. > > *Proof*. Let $m_s(E) = m(s + E)$ and $m^r(E) = m(rE)$. Assume that $r \ne 0$. Suppose that $E$ is $m_s$ or $m^r$ measurable, then since translation and dilation preserves unions and complements, $m_s$ and $m^r$ are both [[Measure Space|measures]]. Moreover, translation and dilation also preserve the elementary family of h-intervals $\ce$. > > For any bounded h-interval $(a, b]$, > $ > \begin{align*} > m_s((a, b]) &= m((a + s, b + s]) \\ > &= (b + s) - (a + s) \\ > &= b - a = m((a, b]) \\ > m^r((a, b]) &= m(r(a, b]) \\ > &= \begin{cases} > m((ra, rb]) & r \ge 0 \\ > m((rb, ra]) &r < 0 > \end{cases} \\ > &= \begin{cases} > rb - ra & r \ge 0 \\ > ra - rb &r < 0 > \end{cases} \\ > &= |r|(b - a) = |r|m((a, b]) > \end{align*} > $ > and for any unbounded interval, $(-\infty, b]$ and $(a, \infty)$, > $ > \begin{align*} > m_s((-\infty, b]) &= m((-\infty, b + s]) \\ > &= \infty = m((-\infty, b]) \\ > m_s((a, \infty)) &= m((a + s, \infty)) \\ > &= \infty = m((a, \infty)) \\ > m^r((-\infty, b]) &= \begin{cases} > m((-\infty, rb]) & r > 0 \\ > m([rb, \infty)) & r < 0 > \end{cases}\\ > &= \begin{cases} > \infty & r \ne 0 \\ > 0 & r = 0 > \end{cases} \\ > &= |r|m((-\infty, b])\\ > m^r((a, \infty)) &= \begin{cases} > m((ra, \infty)) & r > 0 \\ > m((-\infty, ra)) & r < 0 > \end{cases}\\ > &= \begin{cases} > \infty & r \ne 0 \\ > 0 & r = 0 > \end{cases} \\ > &= |r|m((a, \infty)) > \end{align*} > $ > and for $\real$, $m_s(\real) = m(\real)$ and $m^r(\real) = m(\real) = |r|m(\real)$. Therefore we have $m_s|_\ce = m|_\ce$, $m^r|_\ce = |r|m|_\ce$, where $\ce$ is the [[Elementary Family|elementary family]] of all h-intervals. > > Since $m_s, m^r$ are both measures, for any FDU $E = \bigcup_{i = 1}^{n}E_i$ of h-intervals, > $ > \begin{align*} > m_s(E) = \sum_{i = 1}^{n}m_s(E_i) &= \sum_{i = 1}^{n}m(E_i) \\ > m^r(E) = \sum_{i = 1}^{n}m^r(E_i) &= \sum_{i = 1}^{n}|r|m(E_i) > \end{align*} > $ > and $m_s|_\alg = m|_\alg$, $m^r|_\alg = |r|m|_\alg$, where $\alg$ is the [[Algebra|algebra]] consisting of FDUs of $\ce$. This means that $m_s$ and $m^r$ are extensions of $m|_\alg$ and $|r|m|_\alg$ on $\cb_\real = \cm(\alg)$, respectively. Since $m$ is $\sigma$-finite, by [[Carathéodory's Extension Theorem]], $m_s|_{\cb_\real} = m|_{\cb_\real}$, $m^r|_{\cb_\real} = |r|m|_{\cb_\real}$. > > Let $E \in \cb_\real$, then $E = V \setminus N$ where $V = \bigcap_{n \in \nat}U_n \in G_\delta$ and $m(N) = 0$. Then > $ > \begin{align*} > s + E &= (s + V) \setminus (s + N) \\ > &= \paren{\bigcap_{n \in \nat}s + U_n} \setminus (s + N) > \end{align*} > $ > Since each $U_n$ is [[Open Set|open]], each $s + U_n$ is also open and $\bigcap_{n \in \nat}s + U_n \in G_\delta$. As $N \in \cl$ is a [[Null Set|null set]], > $ > \forall \varepsilon > 0, \exists U \text{ open}: U \supset N, \mu(U) < \varepsilon > $ > and > $ > s + U \supset s + N, \mu(s + U) < \varepsilon > $ > meaning that $s + N \in \cl$ is also a null set. Therefore $s + E \in \cl$. > > On the other hand, > $ > \begin{align*} > rE &= (rV) \setminus (rN) \\ > &= \paren{\bigcap_{n \in \nat} rU_n} \setminus (rN) > \end{align*} > $ > If $r \ne 0$, then as each $U_n$ is open, so is each $rU_n$. As $N \in \cl$ is a null set, > $ > \forall \varepsilon > 0, \exists U: U, m(U) < \varepsilon/|r| > $ > and > $ > rU \supset rN, m(rU) = |r|m(U) < \varepsilon > $ > meaning that $rN \in \cl$ is also a null set. If $r = 0$, then $rE = \bracs{0} \in \cb_\real \subset \cl$ is also a null set. Therefore $rE \in \cl$. > > As $s + E, rE \in \cl$ for any $E \in \cl$, we have > $ > m(s + E) = m(E) \quad m(rE) = |r|m(E) \quad \forall E \in \cl > $ > [!theorem] > > Let $\mathcal{C}$ be a family of open balls in $\real^n$, and $U = \bigcup_{B \in \mathcal C}B$. Then for any $c < m(U)$, there exists disjoint $\bracs{B_j}_1^k$ such that $\sum_{j \in [k]}m(B_k) > 3^{-n}c$. > > *Proof*. By the inner regularity of the Lebesgue measure, there exists compact $K \subset U$ with $m(K) > c$, which comes with a finite subcover $\bracs{A_j}_1^m \supset K$. Let $B_1$ be of maximal radius, and for any $j > 1$, let $B_j \in \bracs{A_j}_1^m$ a ball disjoint from $\bigcup_{i < j}B_i$ of maximal radius. This yields a family $\bracs{B_j}_1^k \subset \bracs{A_j}_1^m$. > > Suppose that $A_i \not\in \bracs{B_j}_1^k$, then there exists $j$ such that $A_i \cap B_j \ne \emptyset$. Choose the smallest of such $j$, then $r(A_i) \le 3r(B_j)$. Let $B_j^*$ be the ball centred at the centre of $B_j$ with thrice the radius, then $A_i \subset B^*_j$. This allows covering $K$ with the enlarged balls, hence > $ > c < m(K) \le m\paren{\bigcup_{j \in [k]}B_j^*} \le \sum_{j \in [k]}m(B_j^*) = \sum_{j \in [k]}3^nm(B_j) > $