Today I Learned

February 25th

Today I learned about the Lesbegue measure. The basic idea is that we want to create a Haar (translation-invariant and outer Radon) measure which communicates niecly with our basis elements. To start off, we define $\lambda((a,b))=b-a$ to be our measure on the basis elements of our topology.

Next we extend this to all open sets. This is not actually trivial because even though we can write\[U=\bigcup_\alpha(a_\alpha,b_\alpha)\]for some collection of basis elements $(a_\alpha,b_\alpha),$ these basis elements might overlap. The trick is that our basis elements are totally ordered by containment, so we can intuitively choose the "largest'' basis elements to cover $U$ and not worry about overlap. Making this rigorous is somewhat annoying, but one way to do this is to write\[\mathcal B=\{(a,b):(a,b)\subseteq U\},\]and then $\mathcal B$ is a partially ordered set under $\subseteq.$ Each ascending chain has an upper bound (say, $\RR$), so each $B\in\mathcal B$ is contained in some maximal element of $\mathcal B.$ So we can define a function $\varphi:\mathcal B\to\mathcal B$ taking each basis element to its maximal and define\[\mu(U)=\sum_{B\in\varphi(\mathcal B)}\lambda(B).\]This is disgusting, but it provides the idea of what's to come.

We notice now that we want an outer Radon measure, which requires being outer regular. So it would be lovely if we could just define\[\mu(S)=\inf_{\text{open }U\supseteq S}\mu(U)\]to get outer regular for free. However, this forces us to take $\inf$ over the $\varphi$ function from earlier which we already don't understand very well. The technical trick to avoid $\varphi,$ then, is to let the $\inf$ check for overlaps for us. We take the following definition.

Note now that overlaps between the $(a_\bullet,b_\bullet)$ are entirely permitted, but the $\inf$ will tell us that overlapping is not minimal.

It remains to check that $\mu$ is actually a Haar measure. We will show it is an outer measure today and move towards showing Haar. The easiest part of this showing that $\mu$ is translation-invariant, so we get that out of the way. Indeed, for set $S$ and constant $c\in\RR,$ we have to show that $\mu(S)=\mu(c+S),$ or\[\inf\left\{\sum_{k=0}^\infty(b_k-a_k):S\subseteq\bigcup_{k=0}^\infty(a_k,b_k)\right\}=\inf\left\{\sum_{k=0}^\infty(b_k-a_k):c+S\subseteq\bigcup_{k=0}^\infty(a_k,b_k)\right\}.\]Now, for any collection $\{(a_k,b_k)\}_{k=0}^\infty$ covering $S,$ we note $\{(a_k+c,b_k+c)\}_{k=0}^\infty$ will cover $c+S$ while preserving all of the lengths $b_k-a_k=(b_k+c)-(a_k+c).$ It follows that any bound $M$ occuring in the left-hand set will also appear in the right-hand set, so\[\mu(S)\ge\mu(c+S).\]Taking $S\mapsto c+S$ and $c\mapsto-c$ implies $\mu(c+S)\ge\mu(S),$ so $\mu(S)=\mu(c+S),$ as desired.

Before continuing, we establish a lemma. We show that if $A\subseteq B,$ then $\mu(A)\le\mu(B).$ Expanding, we have to show that\[\inf\left\{\sum_{k=0}^\infty(b_k-a_k):A\subseteq\bigcup_{k=0}^\infty(a_k,b_k)\right\}\le\inf\left\{\sum_{k=0}^\infty(b_k-a_k):B\subseteq\bigcup_{k=0}^\infty(a_k,b_k)\right\}.\]Well, we note that any collection $\{(a_k,b_k)\}_{k=1}^\infty$ covering $B$ will also cover $A,$ so any bound $M$ appearing in the right-hand set also appears in the left-hand set. It follows $\mu(A)\le\mu(B).$

We now show that $\mu$ is outer regular. That is, we have to show\[\mu(S)=\inf_{\text{open }U\supseteq S}\mu(U).\]Quickly, we note that $U\supseteq S$ directly implies $\mu(U)\ge\mu(S),$ so we immediately know\[\mu(S)\le\inf_{\text{open }U\supseteq S}\mu(U)\]from the lemma. To get the other direction, we expanding out the definition of $\mu,$ so we have to show\[\inf\left\{\sum_{k=0}^\infty(b_k-a_k):S\subseteq\bigcup_{k=0}^\infty(a_k,b_k)\right\}\stackrel?\ge\inf_{\text{open }U\supseteq S}\left\{\sum_{k=0}^\infty(b_k-a_k):U\subseteq\bigcup_{k=0}^\infty(a_k,b_k)\right\}.\]Now, note that any collection $\{(a_k,b_k)\}$ covering $S$ on the left-hand side will union to an open set that we can give to the right-hand side, with the same cover. So any bound $M$ appearing in the left-hand set will also appear in the right-hand set, so\[\mu(S)\ge\inf_{\text{open }U\supseteq S}\mu(U).\]Combining the two inequalities gets what we want.

Only now do we try to check that $\mu$ is a measure. Take $\{S_k\}_{k=0}^\infty$ sets so that we want to show\[\sum_{k=0}^\infty\mu(S_k)\stackrel?=\mu\left(\bigcup_{k=0}^\infty S_k\right).\]Expanding, we want to know\[\sum_{k=0}^\infty\inf\left\{\sum_{\ell=0}^\infty(b_\ell-a_\ell):S_k\subseteq\bigcup_{\ell=0}^\infty(a_\ell,b_\ell)\right\}\stackrel?=\inf\left\{\sum_{\ell=0}^\infty(b_\ell-a_\ell):\bigcup_{k=0}^\infty S_k\subseteq\bigcup_{\ell=0}^\infty(a_\ell,b_\ell)\right\}.\]The outer sum on the left-hand side is unnatural, so we write\[\inf\left\{\sum_{k,\ell=0}^\infty(b_{k,\ell}-a_{k,\ell}):S_k\subseteq\bigcup_{\ell=0}^\infty(a_{k,\ell},b_{k,\ell})\right\}\stackrel?=\inf\left\{\sum_{\ell=0}^\infty(b_\ell-a_\ell):\bigcup_{k=0}^\infty S_k\subseteq\bigcup_{\ell=0}^\infty(a_\ell,b_\ell)\right\}.\]Now, in one direction, note that for any bound $M$ in the left-hand side, we can associate it with a countable (!) collection $(a_{k,\ell},b_{k,\ell})$ of basis sets. This collection covers each of the $S_\bullet,$ so it covers their union, so this collection will also appear in the right-hand set. Thus, the bound $M$ also appears in the right-hand side. It follows\[\sum_{k=0}^\infty\mu(S_k)\ge\mu\left(\bigcup_{k=0}^\infty S_k\right).\]This is enough for $\mu$ to be an outer measure, but it remains to check the $\le$ inequality. This is because it is actually impossible to show this for arbitrary sets $S_\bullet,$ even though we haven't used the disjoint condition. I don't currently know enough about measurable sets to be able to fix this.