# The Unapologetic Mathematician

## Lebesgue Measure and Affine Transformations

This didn’t make it up yesterday, due to some preoccupation on my part. Playing catch-up…

The thing that makes Lebesgue measure really special is the way that every part of the real line “looks like” every other part, and the same is true as we scale the line up or down. Once we specify the measure of one interval then the measure of every other interval — and every Lebesgue measurable set — is completely determined. But first, let’s show that this nice behavior actually holds.

Let $T$ be an affine transformation of the real line, defined by $T(x)=\alpha x+\beta$ for some real $\alpha\neq0$ and any real $\beta$. This is invertible, and the inverse function is affine as well: $T^{-1}(x)=\frac{x}{\alpha}-\frac{\beta}{\alpha}$. Given a subset $E\subseteq\mathbb{R}$, we write $T(E)$ for the image of $E$ under the transformation $T$ $T(E)=\{\alpha x+\beta\vert x\in E\}$. I say that the outer and inner Lebesgue measures are both nicely behaved under the transformation $T$. $\displaystyle\mu^*(T(E))=\lvert\alpha\rvert\mu^*(E)$ $\displaystyle\mu_*(T(E))=\lvert\alpha\rvert\mu_*(E)$

We’ll start by assuming that $\alpha>0$. In the other case, $T$ is the composition of the mapping $x\mapsto-x$ and an affine transformation with a positive multiplier, so we’ll have to come back and show that this reflection preserves our measures.

Let $T(\mathcal{S})$ be the collection of all sets of the form $T(E)$ with $E\in\mathcal{S}$. It should be clear that this is a $\sigma$-ring, and we will show that $T(\mathcal{S})=\mathcal{S}$. Indeed, if $E\in\mathcal{P}$ is a semiclosed interval $\left[a,b\right)$, then $E=T(F)$, where $F$ is the semiclosed interval $\left[\frac{a-\beta}{\alpha},\frac{b-\beta}{\alpha}\right)$. And so $E\in T(\mathcal{S})$, and the whole $\sigma$-ring $\mathcal{S}$ generated by $\mathcal{P}$ must be contained in $T(\mathcal{S})$: $\mathcal{S}\subseteq T(\mathcal{S})$. We can apply the exact same reasoning to $T^{-1}$ to conclude that $\mathcal{S}\subseteq T^{-1}(\mathcal{S})$, and thus $T(\mathcal{S})\subseteq\mathcal{S}$. This shows that $\mathcal{S}=T(\mathcal{S})$.

For every Borel set $E$, we define $\mu_1(E)=\mu(T(E))$ and $\mu_2(E)=\alpha\mu(E)$. These are both measures on $\mathcal{S}$. If $\mathcal{E}=\left[a,b\right)$ is a semiclosed interval, we calculate \displaystyle\begin{aligned}\mu_1(E)&=\mu(T(E))\\&=\mu\left(\left[\alpha a+\beta,\alpha b+\beta\right)\right)\\&=(\alpha b+\beta)-(\alpha a\beta)\\&=\alpha(b-a)\\&=\alpha\mu(E)\\&=\mu_2(E)\end{aligned}

so the two measures agree on $\mathcal{P}$. But our extension theorem tells us that a measure on $\mathcal{S}$ is uniquely determined by its values on $\mathcal{P}$. And thus our assertion holds for Lebesgue measure on Borel sets.

Now if we apply these considerations to $T^{-1}$, we can show that \displaystyle\begin{aligned}\mu^*(T(E))&=\inf\left\{\mu(F)\vert F\in\mathcal{S},E\subseteq F\right\}\\&=\inf\left\{\alpha\mu(T^{-1}(F))\vert T^{-1}(F)\in\mathcal{S},E\subseteq T^{-1}(F)\right\}\\&=\alpha\inf\left\{\mu(G)\vert G\in\mathcal{S},E\subseteq G\right\}\\&=\alpha\mu^*(E)\end{aligned}

Replacing $\inf$ by $\sup$ and $\subseteq$ by $\supseteq$, we also reach our conclusion for inner Lebesgue measure. But we still need to verify that $T$ preserves Lebesgue measurability in the first place. If $E$ is Lebesgue measurable, and $A$ is any set, then \displaystyle\begin{aligned}\mu^*(A\cap T(E))+\mu^*(A\cap T(E)^c)&=\mu^*(T(T^{-1}(A)\cap E))+\mu^*(T(T^{-1}(A)\cap E^c))\\&=\alpha\left(\mu^*(T^{-1}(A)\cap E)+\mu^*(T^{-1}(A)\cap E^c)\right)\\&=\alpha\mu^(T^{-1}(A))\\&=\mu^*(A)\end{aligned}

and thus $T(E)$ is also Lebesgue measurable.

Now let’s go back and consider what happens with the reflection $x\mapsto-x$. This sends the semiclosed interval $\left[a,b\right)$ to the interval $\left(-b,-a\right]$. But this is a Borel set: $\left(-b,-a\right]=\left[-b,-a\right)\cup\{-a\}\setminus\{-b\}$, and the singletons are Borel sets. Thus we see that the reflection sends Borel sets to Borel sets. It should also be clear that it preserves Lebesgue measure, for $(-a)-(-b)=b-a$. If $D$ is a subset of a Borel set of measure zero, then the reflection of $D$ is again such a subset, and so we see that reflection preserves the Lebesgue measurable sets as well as their inner and outer measures.

April 22, 2010 - Posted by | Analysis, Measure Theory