Proof of A Measurable Real Function Has Null Level Sets at Levels Arbitrarily Close to Any Point, for a Finite Measure

Throughout, the maps below given by formulas (the choice map $\sigma$ of Step 1, the map $\tau$ of Step 2 and the sequence of Step 3) are maps by Maps and Relations Given by Formulas (i) and Maps and Relations Given by Formulas (ii), their formulas being predicative (The Graph Reading of a Formula Tree Is Predicative and Agrees with Its Reading, So the Reading Is Predicative (iii)). Let $P=\{(c,e)\in\mathbb{R}\times\mathbb{R}:c<e\}$; for $(c,e)\in P$ we write $[c,e]$ for the closed interval. Closed intervals are Borel (Intervals Are Borel, and Rays, Half-Open Intervals and Rational Intervals Generate the Borel Sigma-Algebra of the Real Line (ii)), so $g^{-1}([c,e])\in\mathcal{F}$, and $\mu(g^{-1}([c,e]))$ is a real number since $\mu$ is finite.

Step 1 (keeping the lighter third). Let $(c,e)\in P$ and $J=[c,e]$, and consider the outer thirds

$$J_{1}=\Bigl[c,\ \frac{2c+e}{3}\Bigr],\qquad J_{2}=\Bigl[\frac{c+2e}{3},\ e\Bigr].$$

Both are closed intervals of positive length contained in $J$, and they are disjoint, as $\frac{2c+e}{3}<\frac{c+2e}{3}$. Hence $g^{-1}(J_{1})$ and $g^{-1}(J_{2})$ are disjoint subsets of $g^{-1}(J)$, and by monotonicity and the rule for differences (Basic Properties of a Measure (iii), Basic Properties of a Measure (iv))

$$\mu\bigl(g^{-1}(J_{2})\bigr)\le\mu\bigl(g^{-1}(J)\setminus g^{-1}(J_{1})\bigr)=\mu\bigl(g^{-1}(J)\bigr)-\mu\bigl(g^{-1}(J_{1})\bigr).$$

Let $\sigma(c,e)\in P$ be the pair of endpoints of $J_{1}$ if $\mu(g^{-1}(J_{1}))\le\mu(g^{-1}(J_{2}))$, and of $J_{2}$ otherwise. Writing $J'$ for the interval so chosen, $J'\subseteq J$ and

$$2\,\mu\bigl(g^{-1}(J')\bigr)\le\mu\bigl(g^{-1}(J_{1})\bigr)+\mu\bigl(g^{-1}(J_{2})\bigr)\le\mu\bigl(g^{-1}(J)\bigr).$$

Step 2 (nested intervals and a null level). Let $(c,e)\in P$. By recursion on $\mathbb{N}$ (Recursion on the Natural Numbers Starting at One (i)) there is a sequence $\bigl((c_{n},e_{n})\bigr)_{n\in\mathbb{N}}$ in $P$ with

$$(c_{1},e_{1})=(c,e),\qquad(c_{n+1},e_{n+1})=\sigma(c_{n},e_{n})\quad(n\in\mathbb{N}).$$

Write $I_{n}=[c_{n},e_{n}]$. By Step 1, $I_{n+1}\subseteq I_{n}$, that is, $c_{n}\le c_{n+1}<e_{n+1}\le e_{n}$, and $2\mu(g^{-1}(I_{n+1}))\le\mu(g^{-1}(I_{n}))$. By induction (Arithmetic and Order of the Natural Numbers (xiii)), $c_{m}\le c_{n}$ and $e_{n}\le e_{m}$ whenever $m\le n$; so $c_{m}\le c_{l}<e_{l}\le e_{n}$ with $l$ the larger of $m$ and $n$, that is,

$$c_{m}\le e_{n}\qquad\text{for all }m,n\in\mathbb{N}.$$

The set $\{c_{j}:j\in\mathbb{N}\}$ is nonempty and bounded above by $e_{1}$, so it has a supremum $t$ (The Real Numbers Form an Ordered Field in Which Every Nonempty Set Bounded Above Has a Supremum (ii), The Real Numbers, with the Natural Numbers, Integers and Rationals Identified with Subsets of the Reals, and Completeness (vii)), and $c_{n}\le t\le e_{n}$ for every $n$, since each $e_{n}$ is an upper bound. Hence

$$\{g=t\}=g^{-1}([t,t])\subseteq g^{-1}(I_{n})\qquad(n\in\mathbb{N}),$$

and $\mu(\{g=t\})\le\mu(g^{-1}(I_{n}))$ for every $n$ by monotonicity. The set $\{\mu(g^{-1}(I_{n})):n\in\mathbb{N}\}$ of real numbers is nonempty and bounded below by $0$, so it has an infimum $\delta\ge0$ (Infima, the Archimedean Property, Density of the Rationals and Rational Approximation from Below in an Ordered Field Whose Nonempty Sets Bounded Above Have Suprema (i)), and $\mu(\{g=t\})\le\delta$. In fact $\delta=0$: if $\delta>0$, then $2\delta$ is not a lower bound, so $\mu(g^{-1}(I_{n}))<2\delta$ for some $n$, and then $\mu(g^{-1}(I_{n+1}))\le\frac{1}{2}\mu(g^{-1}(I_{n}))<\delta$, contradicting the choice of $\delta$. Therefore $\mu(\{g=t\})=0$.

The number $t$ depends only on $(c,e)$; write $t=\tau(c,e)$. We have shown that $\tau:P\to\mathbb{R}$ satisfies, for every $(c,e)\in P$,

$$c\le\tau(c,e)\le e,\qquad\mu\bigl(\{g=\tau(c,e)\}\bigr)=0 .$$

Step 3 ((i)). Let $a<b$. Then $\bigl(\frac{2a+b}{3},\frac{a+2b}{3}\bigr)\in P$, and $t=\tau\bigl(\frac{2a+b}{3},\frac{a+2b}{3}\bigr)$ satisfies $a<\frac{2a+b}{3}\le t\le\frac{a+2b}{3}<b$ and $\mu(\{g=t\})=0$ by Step 2.

Step 4 ((ii)). Let $a\in\mathbb{R}$. For $k\in\mathbb{N}$, read as a real number (Maps Given by Iterated Operations: Partial Iterates, Images of Natural Numbers, Powers and Operations over a Finite Set (ii), with $1$ and sums preserved by The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products (i) and The Image of the Natural Numbers with Zero in a Commutative Ring Respects Zero, One, Sums, Products, Differences, Powers, and Finite Sums and Products (ii)), put

$$t_{k}=\tau\Bigl(a+\frac{1}{3k},\ a+\frac{2}{3k}\Bigr).$$

The sequence $(t_{k})$ is the composite of $\tau$ with the map $k\mapsto\bigl(a+\frac{1}{3k},a+\frac{2}{3k}\bigr)$ from $\mathbb{N}$ to $P$ (Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (ii)). By Step 2, for every $k\in\mathbb{N}$,

$$a<a+\frac{1}{3k}\le t_{k}\le a+\frac{2}{3k}<a+\frac{1}{k},\qquad\mu(\{g=t_{k}\})=0 .$$