Proof of Scaling of Lebesgue Measure and the Lebesgue Integral on $\mathbb{R}^n$

Throughout, $\lambda$ is Lebesgue measure on the real line, $\lambda^{*}$ is Lebesgue outer measure, and $T(x)=cx$ for $x\in\mathbb{R}^n$. Points of $\mathbb{R}^n$ are written in coordinates, $x=(x_1,\dots,x_n)$, and the index $j$ always ranges over $\{1,\dots,n\}$. By claims 4 and 5 of Properties of Natural Number Powers in a Field the real number $\alpha=|c|^{n}$ is nonnegative and nonzero, hence positive, and by claims 2 and 3 there $\alpha^{-1}=(|c|^{-1})^{n}$, the number written $|c|^{-n}$ in the statement.

Step 1: positive multiples of sums in $[0,\infty]$. Let $(t_m)_{m\in\mathbb{N}}$ be a sequence in $[0,\infty]$ and let $\beta$ be a positive real number. Then $\sum_m\beta t_m=\beta\sum_m t_m$, the sums being those of Measure, Measure Space, and Probability Measure. Indeed, suppose first that every $t_m$ is real and that the partial sums $s_M=\sum_{m\le M}t_m$ are bounded above, and let $L$ be their least upper bound. The partial sums of $(\beta t_m)$ are the numbers $\beta s_M$; the number $\beta L$ is an upper bound for them, and if $U$ is any upper bound for them then $\beta^{-1}U$ is an upper bound for the $s_M$, so $L\le\beta^{-1}U$ and hence $\beta L\le U$. Thus $\beta L$ is the least upper bound of the partial sums of $(\beta t_m)$, and both sides equal $\beta L$. In the remaining case either some $t_m$ equals $\infty$, and then $\beta t_m=\infty$, or every $t_m$ is real and the $s_M$ are unbounded above, and then so are the $\beta s_M$, since an upper bound $U$ for the latter yields the upper bound $\beta^{-1}U$ for the former. In both cases each side equals $\infty$.

Step 2: the outer measure scales. Let $E\subseteq\mathbb{R}$ and let $b\ne0$ be real. We show $\lambda^{*}(bE)=|b|\,\lambda^{*}(E)$.

Let $(u_m,v_m)_{m\in\mathbb{N}}$ be open intervals with $u_m\le v_m$ and $E\subseteq\bigcup_m(u_m,v_m)$. Put $(p_m,q_m)=(bu_m,bv_m)$ if $b>0$ and $(p_m,q_m)=(bv_m,bu_m)$ if $b<0$. Multiplying the inequalities $u_m\le v_m$ by $b$, which reverses them exactly when $b<0$, gives $p_m\le q_m$, and in both cases $q_m-p_m=|b|(v_m-u_m)$. If $y\in bE$, say $y=bx$ with $x\in E$, then $u_m<x<v_m$ for some $m$, and multiplying by $b$ gives $p_m<y<q_m$; hence $bE\subseteq\bigcup_m(p_m,q_m)$. By step 1, $\sum_m(q_m-p_m)=|b|\sum_m(v_m-u_m)$, so by Lebesgue Outer Measure on the Real Line,

$$\lambda^{*}(bE)\le|b|\sum_m(v_m-u_m).$$

If $\lambda^{*}(E)=\infty$ then $|b|\lambda^{*}(E)=\infty$ and the inequality $\lambda^{*}(bE)\le|b|\lambda^{*}(E)$ is immediate. Otherwise some covering sequence has a finite sum, so the display makes $\lambda^{*}(bE)$ real, and the display rearranges to $|b|^{-1}\lambda^{*}(bE)\le\sum_m(v_m-u_m)$ for every covering sequence with a finite sum, hence for every covering sequence. Therefore $|b|^{-1}\lambda^{*}(bE)$ is a lower bound for the set whose infimum is $\lambda^{*}(E)$, so $|b|^{-1}\lambda^{*}(bE)\le\lambda^{*}(E)$ and again $\lambda^{*}(bE)\le|b|\lambda^{*}(E)$.

Applying this inequality with $b^{-1}$ in place of $b$ and $bE$ in place of $E$, and using $b^{-1}(bE)=E$ together with $|b^{-1}|=|b|^{-1}$, gives $\lambda^{*}(E)\le|b|^{-1}\lambda^{*}(bE)$, that is, $|b|\lambda^{*}(E)\le\lambda^{*}(bE)$. The two inequalities give the asserted equality.

Step 3: $T$ is Borel measurable, and $cB$ is Borel for Borel $B$. The difference $x-y$ has $j$-th coordinate $x_j-y_j$, its Euclidean norm is the Euclidean distance $d(x,y)$ by claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, and claim 4 of the same lemma bounds each coordinate of a point by its norm; so $|x_j-y_j|\le d(x,y)$ for every $j$. Hence if $d(x^k,x)\to0$ for a sequence $(x^k)_{k\in\mathbb{N}}$ in $\mathbb{R}^n$ then $x^k_j\to x_j$, and therefore $cx^k_j\to cx_j$, for every $j$. So every component of $T$ is sequentially continuous in the sense of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, and claim 3(b) there makes $T$ measurable with respect to the $\sigma$-algebra $\mathcal{B}_n$ of that lemma, which is $\mathcal{B}(\mathbb{R}^n)$ by claim 5 there.

For any $B\subseteq\mathbb{R}^n$ one has $T^{-1}(B)=c^{-1}B$: if $cx\in B$ then $x=c^{-1}(cx)\in c^{-1}B$, and if $x=c^{-1}y$ with $y\in B$ then $cx=y\in B$. Applying this with $c^{-1}$ in place of $c$ shows that $cB$ is the preimage of $B$ under the map $x\mapsto c^{-1}x$, which is measurable by the argument just given; so $cB$ is Borel whenever $B$ is.

Step 4: the one-dimensional case. Let $A\in\mathcal{B}(\mathbb{R})$ and let $b\ne0$ be real. The argument of step 3, applied with $1$ in place of $n$ and $b$ in place of $c$, shows that $bA$ is Borel, where $\mathcal{B}(\mathbb{R}^1)=\mathcal{B}(\mathbb{R})$ by Borel Sigma-Algebra on Euclidean Space. By claim 3 of Existence of Lebesgue Measure on the Real Line, $\lambda$ is the restriction of $\lambda^{*}$ to $\mathcal{B}(\mathbb{R})$, so step 2 gives

$$\lambda(bA)=\lambda^{*}(bA)=|b|\,\lambda^{*}(A)=|b|\,\lambda(A).$$

Step 5: claim 1. Let $\mu$ be the image measure of $\lambda_n$ under $T$, a measure on $\mathcal{B}(\mathbb{R}^n)$ by claim 1 of that lemma; by its definition and step 3, $\mu(B)=\lambda_n(c^{-1}B)$ for every Borel $B$. Let $h:\mathbb{R}^n\to[0,\infty)$ be the constant function with value $\alpha$, which is measurable because $\{x\in\mathbb{R}^n:h(x)>r\}$ is $\mathbb{R}^n$ or $\varnothing$ for every real $r$, and let $\nu$ be the measure with density $h$ with respect to $\mu$, a measure by claim 3 of the same lemma. For Borel $B$ the pointwise product $\mathbf{1}_Bh$ is $\alpha\mathbf{1}_B$, so claim 1 of Linearity and Monotonicity of the Lebesgue Integral and The Integral of an Indicator Function is the Measure of the Set give

$$\nu(B)=\int_{\mathbb{R}^n}\mathbf{1}_Bh\,d\mu=\alpha\int_{\mathbb{R}^n}\mathbf{1}_B\,d\mu=\alpha\,\mu(B).$$

Let $A_1,\dots,A_n$ be Borel subsets of $\mathbb{R}$. A point $x$ satisfies $cx\in A_1\times\dots\times A_n$ exactly when $x_j\in c^{-1}A_j$ for every $j$, so

$$T^{-1}(A_1\times\dots\times A_n)=(c^{-1}A_1)\times\dots\times(c^{-1}A_n),$$

each factor being Borel with $\lambda(c^{-1}A_j)=|c|^{-1}\lambda(A_j)$ by step 4. Hence, by Lebesgue Measure on $\mathbb{R}^n$,

$$\mu(A_1\times\dots\times A_n)=\lambda(c^{-1}A_1)\cdots\lambda(c^{-1}A_n)=\bigl(|c|^{-1}\lambda(A_1)\bigr)\cdots\bigl(|c|^{-1}\lambda(A_n)\bigr).$$

Write $P=\lambda(A_1)\cdots\lambda(A_n)$. If some factor $\lambda(A_j)$ is $0$, then the same is true after multiplication by the positive number $|c|^{-1}$, and both the last product and $P$ are $0$; if no factor is $0$ and some factor is $\infty$, then both are $\infty$; and if every factor is real, an induction on $n$ using claim 1 of Properties of Natural Number Powers in a Field gives $\bigl(|c|^{-1}\lambda(A_1)\bigr)\cdots\bigl(|c|^{-1}\lambda(A_n)\bigr)=(|c|^{-1})^{n}P$. Since $(|c|^{-1})^{n}=\alpha^{-1}$ and since $\alpha^{-1}\cdot0=0$ and $\alpha^{-1}\cdot\infty=\infty$, in all three cases

$$\mu(A_1\times\dots\times A_n)=\alpha^{-1}P,\qquad\text{hence}\qquad\nu(A_1\times\dots\times A_n)=\alpha\alpha^{-1}P=P,$$

the last equality again by inspection of the three cases. These are exactly the values that characterize $\lambda_n$ among measures on $\mathcal{B}(\mathbb{R}^n)$ in Lebesgue Measure on $\mathbb{R}^n$, and only one measure has them; therefore $\nu=\lambda_n$, that is,

$$\alpha\,\lambda_n(c^{-1}B)=\lambda_n(B)\qquad\text{for every }B\in\mathcal{B}(\mathbb{R}^n).$$

Applying this with $cB$ in place of $B$, which is legitimate by step 3, and using $c^{-1}(cB)=B$, gives $\alpha\,\lambda_n(B)=\lambda_n(cB)$. This proves claim 1.

Step 6: claims 2 and 3. Multiplying the last display of step 5 by $\alpha^{-1}$ gives $\mu(B)=\alpha^{-1}\lambda_n(B)$ for every Borel $B$. Let $g:\mathbb{R}^n\to[0,\infty)$ be the constant function with value $\alpha^{-1}$, measurable as $h$ was, and let $\nu_g$ be the measure with density $g$ with respect to $\lambda_n$. The computation of step 5, with $\lambda_n$ and $\alpha^{-1}$ in place of $\mu$ and $\alpha$, gives $\nu_g(B)=\alpha^{-1}\lambda_n(B)$ for every Borel $B$; hence $\mu=\nu_g$.

Let $f:\mathbb{R}^n\to[0,\infty]$ be measurable, which by Lebesgue Integral of a Nonnegative Measurable Function means that $\{y\in\mathbb{R}^n:f(y)>r\}$ is Borel for every real $r$. For every real $r$,

$$\{x\in\mathbb{R}^n:f(T(x))>r\}=T^{-1}\bigl(\{y\in\mathbb{R}^n:f(y)>r\}\bigr),$$

which is Borel by step 3; hence $f\circ T$ is measurable. By claim 2 of Image Measures, Measures with Densities, and Change of Variables, applied to $T$, and then by claim 3 of the same lemma, applied to $\mu=\nu_g$,

$$\int_{\mathbb{R}^n}f(cx)\,d\lambda_n(x)=\int_{\mathbb{R}^n}f\circ T\,d\lambda_n=\int_{\mathbb{R}^n}f\,d\mu=\int_{\mathbb{R}^n}f\,g\,d\lambda_n=\alpha^{-1}\int_{\mathbb{R}^n}f\,d\lambda_n,$$

the last equality by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, since $fg$ is the pointwise product of $f$ with the constant $\alpha^{-1}$. This proves claim 2.

Finally let $f:\mathbb{R}^n\to\mathbb{R}$ be measurable. By claim 2 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line the rays $(r,\infty)$, $r\in\mathbb{R}$, generate $\mathcal{B}(\mathbb{R})$, and $(f\circ T)^{-1}\bigl((r,\infty)\bigr)=T^{-1}\bigl(f^{-1}((r,\infty))\bigr)$ is Borel by step 3; so $f\circ T$ is measurable by claim 2 of Generator Criterion for Measurability. By claim 2 of Image Measures, Measures with Densities, and Change of Variables, $f$ is integrable with respect to $\mu$ if and only if $f\circ T$ is integrable with respect to $\lambda_n$, and in that case the two integrals agree. By claim 3 of the same lemma, applied to $\mu=\nu_g$, $f$ is integrable with respect to $\mu$ if and only if the pointwise product $fg=\alpha^{-1}f$ is integrable with respect to $\lambda_n$, and in that case $\int f\,d\mu=\int\alpha^{-1}f\,d\lambda_n$. By claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied with the factor $\alpha^{-1}$ and then with the factor $\alpha$, the function $\alpha^{-1}f$ is integrable with respect to $\lambda_n$ if and only if $f$ is, and then $\int\alpha^{-1}f\,d\lambda_n=\alpha^{-1}\int f\,d\lambda_n$. Combining the three equivalences proves claim 3.