Proof of The Interior of a Convex Set is Convex and Carries Each of Its Borel Subsets up to a Null Set

Each result cited below is universally quantified over the data in its own statement, and is used here for the dimension $n$ fixed in the statement. Throughout, $\operatorname{int}C$ is open by The Interior is the Largest Open Subset, so by Open Subset of a Metric Space every point of it is the centre of an open ball contained in it, and conversely a point that is the centre of an open ball contained in $C$ belongs to $\operatorname{int}C$ by Interior of a Subset of a Topological Space, that ball being open by Open Ball in a Metric Space is Open.

Claim 1. Let $x,y\in\operatorname{int}C$ and let $t\in\mathbb{R}$ with $0\le t\le1$. Choose positive reals $r,s$ with $B(x,r)\subseteq C$ and $B(y,s)\subseteq C$, and let $\rho$ be the least of $r$ and $s$, positive by claim 9 of Elementary Order Arithmetic in an Ordered Field. Let $u\in B(tx+(1-t)y,\rho)$ and put $w=u-(tx+(1-t)y)$, so that $\lVert w\rVert<\rho$ by claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. The same claim gives $d_{E}(x,x+w)=\lVert w\rVert<\rho\le r$, so $x+w\in B(x,r)\subseteq C$, and likewise $y+w\in B(y,s)\subseteq C$. In the real vector space $\mathbb{R}^{n}$ of Euclidean Space $\mathbb{R}^n$ is a Real Vector Space,

$$t(x+w)+(1-t)(y+w)=tx+(1-t)y+\bigl(t+(1-t)\bigr)w=tx+(1-t)y+w=u ,$$

so $u\in C$ by Segments and Convex Subsets of $\mathbb{R}^n$. Hence $B(tx+(1-t)y,\rho)\subseteq C$ and $tx+(1-t)y\in\operatorname{int}C$. Therefore $\operatorname{int}C$ is convex.

Claim 2. Let $E\in\mathcal{B}(\mathbb{R}^{n})$ with $E\subseteq C$ and put $F=E\setminus\operatorname{int}C$, which belongs to $\mathcal{B}(\mathbb{R}^{n})$ because $\operatorname{int}C$ is open, hence Borel by Product Measures and Lebesgue Measure on ℝⁿ: Standing Notation and Background (iv). Suppose, for contradiction, that $\lambda_{n}(F)\ne0$.

2a. A density point in $F$. By claims 1 and 5 of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$, $F$ is not null. By The Lebesgue Density Theorem in $\mathbb{R}^n$ (i) the set of points of $F$ that are not density points of $F$ is null; were that set all of $F$, then $F$ would be null. Hence there is $x\in F$ which is a density point of $F$.

2b. Exterior points arbitrarily close to $x$. Since $x\in F\subseteq C$ and $x\notin\operatorname{int}C$, no open ball centred at $x$ is contained in $C$; so for every positive real $\delta$ there is $z\in B(x,\delta)$ with $z\notin C$, and $z\ne x$ because $x\in C$.

2c. The reflection. Apply Density Point of an Arbitrary Subset of $\mathbb{R}^n$ (i) with $\alpha=\tfrac{3}{4}$, which is less than $1$: there is a positive real $\rho$ with

$$\tfrac{3}{4}\,\lambda_{n}\bigl(\bar{B}(x,r)\bigr)\le\lambda_{n}^{\ast}\bigl(F\cap\bar{B}(x,r)\bigr)\qquad\text{for every real }r\text{ with }0<r<\rho .$$

Fix one such $r$ and put $A=F\cap\bar{B}(x,r)$, a member of $\mathcal{B}(\mathbb{R}^{n})$ by The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ (i); by claim 1 of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$ the displayed inequality reads $\tfrac{3}{4}\lambda_{n}(\bar{B}(x,r))\le\lambda_{n}(A)$.

Let $z\in\mathbb{R}^{n}\setminus C$ with $z\ne x$, and let $A'=2z-A=\{2z-u:u\in A\}$ in the notation of Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ, where $2z=z+z$. By claim 1 of that lemma $A'\in\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{n}(A')=\lambda_{n}(A)$.

The sets $A$ and $A'$ are disjoint. Indeed, let $u\in A\subseteq C$; in the real vector space $\mathbb{R}^{n}$ one has $z=\tfrac12 u+\bigl(1-\tfrac12\bigr)(2z-u)$, so if $2z-u$ belonged to $C$ then $z\in C$ by Segments and Convex Subsets of $\mathbb{R}^n$, contrary to the choice of $z$. Hence $A'\cap C=\emptyset$, and in particular $A'\cap A=\emptyset$.

Both sets lie in a slightly larger closed ball. For $u\in A$ we have $\lVert u-x\rVert\le r$ by claim 2 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square, and, since $(2z-u)-x=2(z-x)-(u-x)$ in $\mathbb{R}^{n}$, claims 5 and 6 of that lemma give

$$\lVert(2z-u)-x\rVert\le2\lVert z-x\rVert+\lVert u-x\rVert\le r+2\lVert z-x\rVert .$$

Thus $A\cup A'\subseteq\bar{B}(x,r+2\lVert z-x\rVert)$, and claims 1 and 2 of Basic Properties of a Measure give

$$2\,\lambda_{n}(A)=\lambda_{n}(A)+\lambda_{n}(A')=\lambda_{n}(A\cup A')\le\lambda_{n}\bigl(\bar{B}(x,r+2\lVert z-x\rVert)\bigr).$$

2d. The contradiction. Write $\kappa_{n}$ for the constant of The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ (ii); by The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ (iii) the two balls above have measures $\kappa_{n}r^{n}$ and $\kappa_{n}(r+2\lVert z-x\rVert)^{n}$. Combining with step 2c,

$$\tfrac{3}{2}\,\kappa_{n}r^{n}\le2\lambda_{n}(A)\le\kappa_{n}\bigl(r+2\lVert z-x\rVert\bigr)^{n}.$$

Now let $k\in\mathbb{N}$. By step 2b, applied with $\delta$ the positive real $2^{-k}$, there is $z_{k}\notin C$ with $0<\lVert z_{k}-x\rVert<2^{-k}$, whence, $t\mapsto t^{n}$ being nondecreasing on the nonnegative reals by claim 5 of Elementary Arithmetic in an Ordered Field and induction on the exponent (the inductive set being the set of $j\in\mathbb{N}$ for which $0\le a\le b$ implies $a^{j}\le b^{j}$),

$$\tfrac{3}{2}\,\kappa_{n}r^{n}\le\kappa_{n}\,\bigl(r+2\cdot2^{-k}\bigr)^{n}.$$

The sequence $(r+2\cdot2^{-k})_{k\in\mathbb{N}}$ converges to $r$, since $(2^{-k})_{k\in\mathbb{N}}$ converges to $0$ by Series of Nonnegative Real Numbers, Comparison, and the Geometric Series (iv) and by the laws for scalar multiples and sums of Arithmetic of Limits of Real Sequences. By the product law of that theorem and induction on the exponent (the inductive set being the set of $j\in\mathbb{N}$ for which the sequence of $j$th powers converges to $r^{j}$), the sequence $\bigl(\kappa_{n}(r+2\cdot2^{-k})^{n}\bigr)_{k\in\mathbb{N}}$ converges to $\kappa_{n}r^{n}$. Passing to the limit in the last display by Order Properties of Limits of Real Sequences gives $\tfrac{3}{2}\kappa_{n}r^{n}\le\kappa_{n}r^{n}$. But $\kappa_{n}r^{n}$ is positive by The Lebesgue Measure of a Closed Ball in $\mathbb{R}^n$ (iii), so claim 10 of Elementary Order Arithmetic in an Ordered Field yields $\tfrac{3}{2}\le1$, contradicting $1<\tfrac{3}{2}$: indeed $\tfrac{3}{2}=1+2^{-1}$, the number $2^{-1}$ is positive by claims 8 and 7 of that lemma, and adding $1$ to $0<2^{-1}$ by claim 1 there gives $1<1+2^{-1}$.

Therefore $\lambda_{n}(F)=0$, which is claim 2.