Mathematical Proofs
This page collects the written mathematical proofs in Appendices A–D. Measurable-version errata, finite-dimensional calculations and historical proof boundaries have separate pages. A written proof is not final mathematical approval.
Mathematical Proofs
A. From Finite to Countable Additivity: The Essential Intermediate Steps
Let \mu^* be an outer measure on a set \Omega, and define
\mathcal M=\{A\subseteq\Omega:\ \forall E\subseteq\Omega,\quad
\mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A)\}.Subadditivity of outer measure always gives the “\le” direction, so proving measurability requires establishing the reverse inequality. We complete the closure and countable-additivity arguments of Section 5.8 below without assuming beforehand that “the measurable sets form a \sigma-algebra.”
A.1 Complements and Finite Unions
We have \varnothing\in\mathcal M. The definition is symmetric in A and A^c, so A\in\mathcal M implies A^c\in\mathcal M.
Take A,B\in\mathcal M. Use A to split E, then use B to split E\setminus A, obtaining
\mu^*(E)=\mu^*(E\cap A)+\mu^*((E\setminus A)\cap B)
+\mu^*(E\setminus(A\cup B)).The first two pieces together are exactly E\cap(A\cup B). Subadditivity says that the sum of their outer measures is at least the outer measure of their union. Thus
\mu^*(E)\ge\mu^*(E\cap(A\cup B))+\mu^*(E\setminus(A\cup B)).Combining this with the reverse inequality, which always holds, gives A\cup B\in\mathcal M. Complements and finite unions then imply closure under finite intersections and differences. This argument uses only addition of nonnegative extended real numbers; it never subtracts a possibly infinite quantity.
A.2 Countable Unions
First suppose that A_n\in\mathcal M are pairwise disjoint, and write A=\bigcup_nA_n. Finitely many splittings give, for every N,
\mu^*(E)=\sum_{n=1}^N\mu^*(E\cap A_n)
+\mu^*\left(E\setminus\bigcup_{n=1}^NA_n\right)
\ge\sum_{n=1}^N\mu^*(E\cap A_n)+\mu^*(E\setminus A).Letting N\to\infty and then applying subadditivity gives
\mu^*(E)\ge\sum_{n=1}^{\infty}\mu^*(E\cap A_n)+\mu^*(E\setminus A)
\ge\mu^*(E\cap A)+\mu^*(E\setminus A).Hence A\in\mathcal M. A general countable union can be expressed, using
B_1=A_1,\qquad B_n=A_n\setminus\bigcup_{k<n}A_kas a union of pairwise disjoint measurable sets. Thus \mathcal M is indeed a \sigma-algebra.
A.3 Why the Restricted Measure Is Countably Additive
Again suppose that A_n are pairwise disjoint. Substituting E=A in the finite splitting above gives
\mu^*(A)\ge\sum_{n=1}^N\mu^*(A_n).Taking the supremum gives the “\ge” direction; subadditivity gives “\le.” Therefore
\mu^*\left(\bigcup_nA_n\right)=\sum_n\mu^*(A_n).Thus \mu=\mu^*|_{\mathcal M} is a measure. We must distinguish “an outer measure is subadditive on all sets” from “its restriction to measurable sets is countably additive”: the latter is precisely the additional structure provided by the splitting criterion.
B. When an Open Cover Is Split, What Happens to the Cutting Point?
Chapter 5 defines m^* through covers by open intervals. To prove that H=(-\infty,a] is measurable, we cannot simply split (u,v) at a into two open intervals and ignore the cutting point.
Fix E and \varepsilon>0. If m^*(E)=\infty, subadditivity already forces
m^*(E\cap H)+m^*(E\setminus H)=\infty,so equality holds. If m^*(E)<\infty, choose an open-interval cover E\subseteq\bigcup_jI_j whose total length is less than m^*(E)+\varepsilon.
Intervals that do not cross a need no splitting. For an interval I_j=(u_j,v_j) that crosses the cutting point, use
(u_j,a+\delta_j),\qquad(a,v_j)to cover its portions in H and H^c, respectively, where \delta_j>0 and \sum_j\delta_j<\varepsilon. Extending the left portion slightly to the right ensures that the cutting point itself is retained; the total length increases by at most \varepsilon. Hence
m^*(E\cap H)+m^*(E\setminus H)
\le\sum_j|I_j|+\sum_j\delta_j
<m^*(E)+2\varepsilon.Letting \varepsilon\downarrow0 and applying subadditivity proves measurability. Intervals with an endpoint equal to a can be handled on the side to which they already belong; a left-hand piece containing a still uses the same slight enlargement.
Closure of the \sigma-algebra makes (-\infty,a)=\bigcup_n(-\infty,a-1/n] measurable, and consequently all open intervals are measurable. Every open set is a countable union of open intervals with rational endpoints, so all Borel sets are measurable. This supplies the complete bridge from interval length to a Borel measure.
C. Product Measure and Tonelli: From Rectangles to General Sets
The phrase “by a monotone-class argument” in Section 6.12 carries substantial content. We expand it here on \sigma-finite spaces (X,\mathcal A,\mu) and (Y,\mathcal B,\nu). We consider only the product \sigma-algebra \mathcal A\otimes\mathcal B; completing it requires separate treatment of exceptions involving sections of null sets.
C.1 Measurable Sections Before Integration
For E\subseteq X\times Y, write E_x=\{y:(x,y)\in E\}. The family of sets for which “every E_x\in\mathcal B” is closed under complements and countable unions and contains all rectangles; it therefore contains \mathcal A\otimes\mathcal B. The same applies to sections in the other direction.
C.2 The Set Identity for Finite Measures
First assume \mu(X),\nu(Y)<\infty. Let \mathcal D consist of all product-measurable sets for which
x\mapsto\nu(E_x),\quad y\mapsto\mu(E^y)are measurable and
\int_X\nu(E_x)\,d\mu(x)=\int_Y\mu(E^y)\,d\nu(y)holds. A rectangle A\times B belongs to \mathcal D, with both sides equal to \mu(A)\nu(B). The whole space also belongs to this family.
Finite total measure makes complements legitimate: a section's measure becomes the constant \nu(Y) minus the original function, and the integral identity follows by subtracting both sides from \mu(X)\nu(Y). For pairwise disjoint E_n\in\mathcal D, the section measures are sums of nonnegative functions; the monotone convergence theorem gives measurability and the integral identity. Thus \mathcal D is a Dynkin system.
Rectangles are closed under finite intersections. By the \pi-\lambda theorem, \mathcal D contains the entire product \sigma-algebra they generate. The \pi-\lambda theorem invoked here is a set-theoretic tool, not Tonelli's theorem, so there is no circular use of the result being proved. A full proof of this set-theoretic tool is given in the π–λ companion.
C.3 The General \sigma-Finite Case and the Construction of the Measure
Choose X_k\uparrow X and Y_k\uparrow Y, each with finite measure. Apply the preceding result to E\cap(X_k\times Y_k) and then use monotone convergence. This gives measurable section functions for general sets and equality of the two iterated integrals. Define
\rho(E)=\int_X\nu(E_x)\,d\mu(x),\qquad E\in\mathcal A\otimes\mathcal B.We have \rho(\varnothing)=0. If E_n are pairwise disjoint, their respective sections are also pairwise disjoint, so
\rho\left(\bigcup_nE_n\right)
=\int_X\sum_n\nu((E_n)_x)\,d\mu
=\sum_n\rho(E_n).Thus \rho is a measure, and on rectangles \rho(A\times B)=\mu(A)\nu(B), with the convention 0\cdot\infty=0. Uniqueness on finite rectangles follows from \pi-\lambda uniqueness; a countable cover then gives uniqueness under \sigma-finiteness. This constructs \mu\otimes\nu without first assuming its existence and using it to prove its own properties.
C.4 From Sets to Functions
The set identity is Tonelli's formula for indicator functions. Finite linear combinations give the formula for nonnegative simple functions. For nonnegative product-measurable f, choose s_n\uparrow f and apply monotone convergence on the product space and in both section directions to obtain
\int f\,d(\mu\otimes\nu)
=\int_X\left(\int_Yf(x,y)\,d\nu(y)\right)d\mu(x)
=\int_Y\left(\int_Xf(x,y)\,d\mu(x)\right)d\nu(y).Each side may equal +\infty. For signed f, first apply Tonelli to |f|. If the product integral is finite, almost all sections are absolutely integrable. Apply the formula separately to f^+ and f^- and subtract to obtain Fubini's theorem. Absolute integrability ensures that we have not silently performed \infty-\infty.
D. Why a Finite Second Moment Suffices: The Characteristic-Function Remainder
In Chapter 3's characteristic-function explanation of the central limit theorem, the crucial step is not formally expanding the exponential, but justifying taking expectations of its second-order expansion. Assume \mathbb EY=0 and \mathbb EY^2=\sigma^2<\infty. Set
r(u)=e^{iu}-1-iu+\frac{u^2}{2}.The second-order Taylor expansion gives r(u)/u^2\to0. There is also an absolute constant C such that |r(u)|\le Cu^2 for all real arguments: for |u|\le1, use the Taylor remainder; for |u|>1, use 2+|u|+u^2/2\le(7/2)u^2.
For t\ne0,
\frac{r(tY)}{t^2}\longrightarrow0,\qquad
\left|\frac{r(tY)}{t^2}\right|\le CY^2.The dominated convergence theorem therefore gives
\varphi_Y(t)=1-\frac{\sigma^2t^2}{2}+o(t^2).If \sigma>0, independent and identically distributed Y_j satisfy
\varphi_{(Y_1+\cdots+Y_n)/(\sigma\sqrt n)}(t)
=\left(1-\frac{t^2}{2n}+o(n^{-1})\right)^n\longrightarrow e^{-t^2/2}.Only after applying Lévy's continuity theorem do we obtain convergence in distribution. No third moment is needed: formally putting the random variable into a third-order remainder and assuming integrability would silently strengthen the theorem's hypotheses. Dominated convergence, established in Chapter 6, supplies exactly this missing logical link in Chapter 3.