# Lebesgue Measurable But Not Borel

/*not*a Borel set. Such a set exists because the Lebesgue measure is the

*completion*of the Borel measure. (The collection $\mathscr{B}$ of Borel sets is generated by the open sets, whereas the set of Lebesgue measurable sets $\mathscr{L}$ is generated by both the open sets

*and*zero sets.) In short, $\mathscr{B}\subset \mathscr{L}$, where the containment is a proper one.

To produce a set in $\mathscr{L}\smallsetminus \mathscr{B}$, we'll assume two facts:

- Every set in $\mathscr{L}$ with positive measure contains a non (Lebesgue) measurable subset.
- 97.3% of all counterexamples in real analysis involve the Cantor set.

*really*a fact, but it may not surprise you that the Cantor set is central to today's discussion. In summary, we will define a homeomorphism (a continuous function with a continuous inverse) from $[0,1]$ to $[0,2]$ which will map a (sub)set (of the Cantor set) of measure 0 to a set of measure 1. By fact #1, this set of measure 1 contains a non-measurable subset, say $N$. And the preimage of $N$ will be Lebesgue measurable but will not be a Borel set. We'll fill in the details below, and while we do, keep in mind that we must work with a

**homeomorphism**- a merely continuous function just won't do. And by playing with the Cantor set, we'll see that

*homeomorphisms*(much less continuous functions!)

*. It's because of this that we can produce a Lebesgue measurable set which is not Borel.*don't always preserve measure

- $f$ is strictly increasing
- since $f'=1$ almost everywhere (recall $c'=0$ almost everywhere)
- $f$ is continuous
- since both $c$ and $x$ are continuous
- $f^{-1}$ exists
- $f$ is 1-1 since it's strictly increasing; it's onto by the Intermediate Value Theorem: since $f(0)=0$, $f(1)=2$ and $f$ is continuous, it assumes all values in between $0$ and $2$!
- $f^{-1}$ is continuous (hence $f$ is a homeomorphism)
- see footnote *

*same length***. This implies $$\mu(f([0,1]\smallsetminus \mathscr{C}))=\mu([0,1]\smallsetminus \mathscr{C})=1.$$ But since $[0,2]= f(\mathscr{C})\sqcup f([0,1]\smallsetminus \mathscr{C})$, we see that $2=\mu([0,2])=\mu(f(\mathscr{C}))+\mu(f([0,1]\smallsetminus \mathscr{C}))=\mu(f(\mathscr{C}))+1$ whence $$\mu(f(\mathscr{C}))=1.$$ From this we deduce that $f(\mathscr{C})\subset[0,2]$ contains a non-measurable subset, say $N$ (see fact #1 in the introduction). And here is where we make our

**Claim: ** $f^{-1}(N)$ is Lebesgue measurable but *not* Borel.

**Lemma:** A strictly increasing function defined on an interval maps Borel sets to Borel sets.

## Proof of Lemma

*Real Analysis*(4ed). Let $f$ be any strictly increasing function defined on some interval. By our analysis above, we know that such a function is a homeomorphism. This fact enables us to show that $f$ maps Borel sets to Borel sets. To do so, it suffices to show that for any continuous function $g$ the set $$\mathscr{A}=\{E:g^{-1}(E) \text{ is Borel} \}$$ is a $\sigma$-algebra containing the open sets. Once we show this, we can conclude $\mathscr{A}$ contains all the Borel sets and therefore, taking $g$ to be $f^{-1}$ (which we know is continuous!), we'll have $(f^{-1})^{-1}(E)=f(E)$ is Borel for any Borel set $E$, which is what we want.

Showing $\mathscr{A}$ is a $\sigma$-algebra (the first two bullets) which contains the open sets (the third bullet) is simple enough (recall that $\mathscr{B}$ denotes the Borel sets):

- If $\{E_i\}\subset\mathscr{A}$ then $f^{-1}(\cup E_i)=\cup f^{-1}(E_i) \in \mathscr{B}$ since $\mathscr{B}$ is a $\sigma$-algebra, hence $\cup E_i\in \mathscr{A}$.
- If $E\subset \mathscr{A}$ then $f^{-1}(E^c)=(f^{-1}(E))^c\in\mathscr{B}$ since $\mathscr{B}$ is a $\sigma$-algebra, hence $E^c\in\mathscr{A}$.
- If $U$ is open, then $f^{-1}(U)$ is open and thus an element of $\mathscr{B}$. Hence $U\in\mathscr{A}$.

## Proof of Claim

*not*Borel! If it were, then since $f$ maps Borel sets to Borel sets by our Lemma, we'd have that $f(f^{-1}(N))=N$ is Borel. But that's impossible since $N$ isn't even measurable! This proves the claim.

*Footnotes*

* *Proof:* Let $h=f^{-1}:[0,2]\to[0,1]$ and suppose $U\subset[0,1]$ is open. Then $[0,1]\smallsetminus U$ is compact and hence closed (and bounded). Since $f$ is continuous, $f([0,1]\smallsetminus U)$ is also closed. But we can rewrite this as
\begin{align*}
f([0,1]\smallsetminus U)&=f([0,1])\smallsetminus f(U)\\
&=[0,2]\smallsetminus f(U)\\
&=[0,2]\smallsetminus h^{-1}(U)
\end{align*}
which allows us to conclude $h^{-1}(U)$ is open.

** *Proof:* This follows simply because $c$ is constant on any interval in $[0,1]\smallsetminus \mathscr{C}$. Indeed for any interval $(a,b)\subset[0,1]\smallsetminus\mathscr{C}$, we have $c(a)=c(b)$ and so
\begin{align*}
\mu((f(a),f(b)))&=f(b)-f(a)\\
&=c(b)+b-c(a)-a\\
&=b-a.
\end{align*}

*References*

- Much of today's discussion is taken from here.

- see also *Real Analysis* (4ed) by Royden, section 2.7, Propositions 21 and 22