Property (1.) in Definition 1.1 is obvious (by monotonicity of measures) and properties (2.) and (3.) are simple consequences of monotone convergence theorems for probability measures.

The “if” direction is clear. To prove the “only if” direction, assume that \(F\) is a degenerated c.d.f., and let \(x_0 = \inf \left\{ x \in \mathbb {R}\; \big| \; F(x) = 1 \right\} \). Then it is straightforward to show by properties of a c.d.f. that \(F\) has the asserted form.

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Fix \(x \in \mathbb {R}\). Note that \(\max (X, Y) \le x\) if and only if both \(X \le x\) and \(Y \le x\). Calculate, using independence,

Induction on \(n\) using 1.7.

Use 1.8 and do a change of variables.

Direct calculations.

Straightforward from the definitions.

Straightforward.

A c.d.f. is always continuous from the right.

Continuity of \(F\) from the left at \(x\) means that for any sequence \((x_n)_{n \in \mathbb {N}}\) increasing to \(x\) (i.e., \(x_n \le x_{n+1} {\lt} x\) for all \(n \in \mathbb {N}\))

or equivalently in terms of measures

But by monotone convergence of measures, we always have

A comparison of these conditions shows that \(F\) is also continuous from the left at \(x\) if and only if \(\mu [\left\{ x \right\} ] = 0\).

Since \(G\) is nondegenerate, there exists some \(x_0 \in \mathbb {R}\) such that \(0 {\lt} G(x_0) {\lt} 1\). Since \(G\) is continuous from the right, for some small \(\delta {\gt} 0\) we have that \(0 {\lt} G(x) {\lt} 1\) for all \(x \in [x_0,x_0+\delta )\). Since the continuity points of \(G\) are dense, from any nonempty open interval we may pick a continuity point. First pick a continuity point \(x_1 \in (x_0,x_0+\delta )\), and then pick another continuity point \(x_2 \in (x_1,x_0+\delta )\).

We must prove that for any \(x \in \mathbb {R}\) we have \(F(x) = G(x)\). But by right-continuity of c.d.f.s, density of \(S\), and coincidence of \(F\) and \(G\) on \(S\), we have

We may first write \(a = A(1) - A(0)\) and \(b = A(0)\). These depend continuously on \(A\), since the evaluations \(A(1)\) and \(A(0)\) do.

The essential claim is that the function

obtained by mapping \(A(x) = a x + b\) to its coefficients \((a,b)\) is a homeomorphism. (The homeomorphism to \(\mathbb {R}^2 = \mathbb {R}\times \mathbb {R}\) follows by combining with the homeomorphism \((a,b) \mapsto (\log a, b)\).)

The continuity of \(\mathrm{cfs}\) follows from Lemma 1.19. For the continuity of the inverse, we must only check that for any \(x \in \mathbb {R}\), its composition with the point evaluation at \(x\) is continuous. But the composition is \((a,b) \mapsto a x + b\), and the continuity is clear.

Calculate and use Lemma 1.19.

The spaces are metrizable, so it suffices to check sequential continuity.

Suppose that \(A_n \to B\) (oriented affine isomophisms) and \(F_n \overset {\mathrm{d}}{\longrightarrow }G\) (c.d.f.s) as \(n \to \infty \). Let \(x \in \mathbb {R}\) be a continuity point of \(B.G\). Let \(\varepsilon {\gt} 0\). Note that \(B^{-1}(x)\) is a continuity point of \(G\). Then there exists a \(\delta {\gt} 0\) such that \(|G(y) - G\big(B^{-1}(x)\big)| {\lt} \frac{\varepsilon }{2}\) when \(|y-B^{-1}(x)|{\lt}\delta \).

By density of continuity points of \(G\), pick continuity points \(y_-,y_+\) such that

Since \(F_n \overset {\mathrm{d}}{\longrightarrow }G\), we have that \(F_n(y_\pm ) \to G(y_\pm )\) as \(n \to \infty \), and in particular there exists some \(N\) such that for \(n \ge N\) we have \(\big| F_n(y_\pm ) - G(y_\pm ) \big| {\lt} \frac{\varepsilon }{4}\) for both \(y_\pm \).

By Lemma 1.21 and \(A_n \to B\), we get that \(A_n^{-1} \to B^{-1}\), and in particular there exists some \(N'\) such that for \(n \ge N'\) we have

For \(n \ge \max \left\{ N,N' \right\} \), we then have

and similarly

which together yield \(\big| (A_n.F_n)(x) - (B.G)(x) \big| {\lt} \varepsilon \).

Since \(x\) was an arbitrary continuity point of \(B.G\) and \(\varepsilon {\gt} 0\) was arbitrary, this proves that \(A_n.F \overset {\mathrm{d}}{\longrightarrow }B.G\).