Image of an increasing function with discontinuities on a dense countable set (e.g. rationales): Cumulative distribution functions have Borel images

Question

When discussing this question, we made interesting observations on increasing (i.e., nondecreasing) functions with discontinuities on rationals. Let $f: [0,1] \to [0,1]$ be an increasing right-continuous function with discontinuities on rationals whose jump values sum to $1$ (see here for examples). It can be shown that this function is strictly (monotonically) increasing, which next implies that its image is a null uncountable set (see this answer). The image of $f$ is Lebesgue measurable, and it seems it is also Borel. This motivates me to ask this question:

Let $f: [0,1] \to \mathbb R$ be an increasing function with discontinuities on a [countable] set that is dense in $[0,1]$. When is the image of $f$ a Borel uncountable set and when is a null set?

I guess that the first holds under mild conditions. I am also wondering how the result changes if $f(0)$ and $f(1)$ are not necessarily finite numbers and the function can tend to infinity at the boundaries, which allows us to extend the result to the case that $\text{dom}(f)$ is not necessarily a closed and bounded interval (I think this case can be handled by applying a strictly increasing continuous transform to $f$ that maps its image to a bounded interval).

As the set of discontinuities of any increasing function is countable [4], "countable" can be removed from the above statement.

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Remark 1. Note such a result is not trivial. Though the image of any injective continuous function with a Borel domain is also Borel [5], but the image of a function with a Borel domain is not necessarily Borel even if it is measurable [6, 7] (any cdf is Borel measurable [8], which also follows from more general result given in [9]).

Remark 2. A somehow counterintuitive implication is that if the support of a discrete random variable is a countably infinite set that can be ordered with respect to the usual order, then the image of its cdf is a countable set; however, if the support of the random variable is a densely ordered countable set, the image becomes an uncountable set, which is similarly null and Borel.

Remark 3. Another interesting implication is that the image of any cumulative distribution function (cdf) is Borel. It follows from the Lebesgue's decomposition theorem that implies the cdf $F_X$ of any random variable $X$ can be written as

$$F_X(x)=\alpha F_{C}(x)+(1-\alpha ) F_D(x)$$

for some $\alpha \in [0,1]$ where $F_{C}(x)$ and $F_D(x)$ are cdfs of some continuous and discrete distributions, respectively.

Remark 4. Another implication is an equivalent definition for a discrete random variable: A random variable is discrete iff the image of its cdf is a null set.

Answer 1

Setup: Suppose $f:[0,1]\rightarrow\mathbb{R}$ is nondecreasing. Let $Q\subseteq[0,1]$ be the set of all points at which it is discontinuous. Suppose $Q$ is dense in $[0,1]$.

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For each $x \in (0,1)$ define \begin{align} f(x^+) &= \lim_{\delta\searrow 0} f(x+\delta)\\ f(x^-) &= \lim_{\delta \searrow 0} f(x-\delta) \end{align} where the limits exist and are finite since $f$ is nondecreasing and bounded above by $f(1)$ and below by $f(0)$. Formally define $f(0^-)=f(0)$ and $f(1^+)=1$.

Claim 1: For all $x\in [0,1]$ we have $$ f(x^-)\leq f(x) \leq f(x^+)$$ and if $x \in Q$ then either $f(x^-)<f(x)$ or $f(x)<f(x^+)$.

Proof: The first inequality follows because $f$ is nondecreasing. If $x\in Q$ then $f$ has a discontinuity at $x$, which means either $f(x^-)\neq f(x)$ or $f(x^+)\neq f(x)$. $\Box$

Claim 2: $f:[0,1]\rightarrow\mathbb{R}$ is strictly increasing.

Proof: Fix $x,y \in [0,1]$ with $x<y$. Since $Q$ is dense in $[0,1]$, there is a $q\in Q$ such that $x<q<y$ and so $$ f(x)\leq f(q^-)\overset{(a)}{\leq} f(q)\overset{(b)}{\leq} f(q^+)\leq f(y)$$ and at least one of the inequalities must be strict, so $f(x)<f(y)$. $\Box$

Claim 3: $f([0,1])$ is uncountably infinite.

Proof: Define $A = f([0,1])$. The function $f:[0,1]\rightarrow\mathbb{R}$ can also be viewed as a function $f:[0,1]\rightarrow A$. We already know $f$ is strictly increasing and hence injective, so $|[0,1]|\leq |A|$. $\Box$

We know that nondecreasing functions can have at most countably many discontinuities, so $Q$ must be countably infinite (it cannot be a finite set since it is dense in $[0,1]$). So we can list the elements of $Q$ as $$ Q = \cup_{i=1}^{\infty} q_i$$ Define $Z$ as the following countably infinite set: $$ Z= \cup_{i=1}^{\infty} \{f(q_i)\}$$ For each $i \in \{1, 2, 3, ...\}$ define $V_i$ and $\overline{V}_i$ as the following open and closed intervals, respectively \begin{align} V_i&=(f(q_i^-), f(q_i^+))\\ \overline{V}_i &=[f(q_i^-), f(q_i^+)] \end{align} Define $p_i=f(q_i^+)-f(q_i^-)$ as the (positive) size of interval $V_i$.

Claim 4: $f([0,1])\setminus Z$ is disjoint from $\cup_{i=1}^{\infty}V_i$.

Proof: Suppose not. Then there is a $t \in [0,1]$ (with $t\notin Q$) and an $i \in \{1, 2, 3, ...\}$ such that $f(t)\in V_i$. Thus $$ f(q_i^-)< f(t) < f(q_i^+)$$ Since $t\neq q_i$ we must have either $t<q_i$ (in which case $f(t)\leq f(q_i^-)$, a contradiction) or $t>q_i$ (in which case $f(t)\geq f(q_i^+)$, again a contradiction). $\Box$

Note that $f(0)<f(1)$ and $$Z \subseteq [f(0), f(1)]$$ $$f([0,1])\subseteq[f(0),f(1)]$$

Claim 5: The sets $\overline{V}_i$ are disjoint subsets of $[f(0), f(1)]$.

Proof: Fix $i\neq j$. WLOG assume $q_i<q_j$. Since $Q$ is dense in $[0,1]$, there is a $q \in Q$ such that $q_i<q<q_j$ and so $$f(q_i^+) \leq f(q^-) \overset{(a)}{\leq} f(q)\overset{(b)}{\leq} f(q^+)\leq f(q_j^-)$$ where either (a) or (b) is strict. So the right endpoint of interval $\overline{V}_i$ is strictly less than the left endpoint of interval $\overline{V}_j$. $\Box$

Claim 6: $f([0,1])$ is a Borel set. Specifically $$ f([0,1]) = Z \cup E \cup \left([f(0), f(1)]\setminus \cup_{i=1}^{\infty} \overline{V}_i\right)$$ where $$ \overline{V}_i= [f(q_i^-), f(q_i^+)]$$ and $E$ is the set of endpoints of these intervals that are in $f([0,1])$, that is, $E$ is the countable subset of $\cup_{i=1}^{\infty} \{f(q_i^-), f(q_i^+)\}$ that is in $f([0,1])$.

Proof: Claim 4 already proves $\subseteq$. It suffices to show $\supseteq$.

Suppose $y \in Z \cup E \left([f(0), f(1)]\setminus \cup_{i=1}^{\infty} \overline{V}_i\right)$. We want to show $y \in f([0,1])$. If $y \in Z$ then $y=f(q)$ for some $q\in Q$ and we are done. If $y \in E$ then (by definition of $E$) $y \in f([0,1])$ and we are done. If $y=f(0)$ or $y=f(1)$ we are also done. Suppose $y \notin Z\cup E$ and $y \neq f(0)$, $y\neq f(1)$. Then $y \in (f(0), f(1))$, but $y$ is not in $\cup_{i=1}^{\infty} \overline{V}_i$. We want to show existence of $t \in [0,1]$ such that $f(t)=y$.

Define $$ z = \inf\{f(x) : f(x)>y, x \in [0,1]\}$$ This is an infimum over a nonempty set because $f(1)>y$. It is clear that $$ y\leq z \leq f(1) \quad (Eq. 1) $$ By definition of $z$ and the fact $f$ is strictly increasing, there is a nonincreasing sequence of real numbers $x_k\in [0,1]$ that satisfy $f(x_k)>y$ for all $k$, and $$ \lim_{k\rightarrow\infty} f(x_k) = z$$ Define $$ t = \lim_{k\rightarrow \infty} x_k$$ where the limit exists because $\{x_k\}$ is nonincreasing in $k$. Then $0\leq t\leq 1$. Then $$ f(t)\leq f(t^+) = \lim_{k\rightarrow\infty} f(x_k)=z$$ and since (Eq. 1) ensures $z\geq y$, we know $f(t^+)\geq y$.

Case 1: Suppose $y\geq f(t^-)$. Then $$f(t^-)\leq y\leq f(t^+)$$ If $t \notin Q$ then $f$ is continuous at $t$ and so $$f(t^-)=f(t)=f(t^+)$$ so that $y=f(t)$ and we are done. On the other hand, if $t \in Q$ then $t=q_m$ for some $m$ and we have $$ f(q_m^-)\leq y \leq f(q_m^+)$$ and so $y \in\overline{V}_m$, a contradiction (so $t \in Q$ cannot occur).

Case 2: Suppose $y< f(t^-)$. Note that $t\neq 0$ because then $y<f(t^-)=f(0)$, but we have assumed $y>f(0)$. Since $y<f(t^-)$, there is an $x \in [0, t)$ such that $$ y<f(x) \leq f(t^-)$$ Since $f(x)>y$ we know by definition of $z$ that $$f(x)\geq z \quad (Eq. 2)$$ On the other hand, since $f$ is strictly increasing, $x<t$ implies $f(x)<f(t)$. We obtain $$ f(t^+)=z\leq f(x) < f(t)\leq f(t^+)$$ where we have used (Eq. 2). This yields the contradiction $f(t^+)<f(t^+)$. So Case 2 cannot occur. $\Box$

Let $\mathcal{B}(\mathbb{R})$ be the standard Borel sigma algebra on $\mathbb{R}$ and let $\mu:\mathcal{B}(\mathbb{R})\rightarrow[0, \infty]$ be the standard Borel measure. Note that $\mu$ maps intervals to their lengths, so $\mu(\overline{V}_i)=p_i$ for all $i \in \{1, 2, 3, ...\}$.

Claim 7: We have $$\mu(f([0,1]))=f(1)-f(0) - \sum_{i=1}^{\infty}p_i$$ In particular, the set $f([0,1])$ has measure 0 if and only if $\sum_{i=1}^{\infty}p_i=f(1)-f(0)$.

Proof: We know $$f([0,1]) = Z \cup E \cup \left([f(0),f(1)]\setminus\cup_{i=1}^{\infty}\overline{V}_i \right) $$ Since $Z$ and $E$ are countable subsets we know $\mu(Z)=\mu(E)=0$. So \begin{align} \mu(f([0,1])) &=\mu([f(0),f(1)]\setminus\cup_{i=1}^{\infty}\overline{V}_i)\\ &= \mu([f(0),f(1)]) - \mu(\cup_{i=1}^{\infty}\overline{V}_i)\\ &=f(1)-f(0)-\sum_{i=1}^{\infty} \underbrace{\mu(\overline{V}_i)}_{p_i} \end{align} where we have used the fact that the sets $\overline{V}_i$ are disjoint (Claim 5). $\Box$

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Examples: Let $Q$ be the set of all rationals in $(0,1)$. Let $\{q_i\}_{i=1}^{\infty}$ be a listing of the elements of $Q$. Consider the functions $f:[0,1]\rightarrow\mathbb{R}$ and $g:[0,1]\rightarrow\mathbb{R}$ defined by

\begin{align} f(x) &= \sum_{i=1}^{\infty} 1_{\{q_i\leq x\}}\frac{1}{2^i} \quad \forall x \in [0,1]\\ g(x) &= x+f(x) \quad \forall x \in [0,1]\\ \end{align} These functions satisfy the setup. Note that $f(0)=g(0)=0$, $f(1)=1$, $g(1)=2$. Then $f([0,1])$ and $g([0,1])$ are both uncountable sets, and $$\mu(f([0,1]))=0 \quad , \quad \mu(g([0,1])) = 1$$

Answer 2

Extension

To extend Claims 6 and 7 in @Michael's answer for $f: (0,1) \to \mathbb R $ (where $f$ is nondecreasing and has discontinuities on dense subset of (0,1)), we can use this representation:

$$(0, 1) =\bigcup_{n=1}^{\infty} \left [\frac{1}{n},1-\frac{1}{n} \right ],$$

which implies

$$f\left((0, 1)\right) =\bigcup_{n=1}^{\infty} f\left(\left [\frac{1}{n},1-\frac{1}{n} \right ]\right). $$

As each $f\left([1/n,1-1/n]\right)$ is a Borel set (Claim 6 in @Michael's answer), it follows that $f\left((0, 1)\right)$ is also Borel.

Moreover, considering that $f\left([1/n,1-1/n]\right), \, n\in\mathbb N$ is an increasing sequence of sets and by Claim 7 in @Michael's answer, we have

$$\mu \left (f((0,1)) \right )=\mu \left (\bigcup_{n=1}^{\infty} f \left ( \left [\frac{1}{n},1-\frac{1}{n} \right ] \right ) \right )= \lim_{n \to \infty} \mu \left (f \left ( \left [\frac{1}{n},1-\frac{1}{n} \right ] \right ) \right ) = \lim_{n \to \infty} \left ( f(1-\frac{1}{n})-f(\frac{1}{n}) - \sum_{i=1}^{\infty} p_i \, I_{[1/n,1-1/n]}(p_i) \right )=f(1^-)-f(0^+) - \sum_{i=1}^{\infty}p_i$$

whenever the right and left limits $f(1^-)$ and $f(0^+)$ are finite.

Other cases where the domain of $f$ is $[0, 1) $ or $(0, 1] $ can be handled similarly. Unbounded domains such as $\mathbb R$, $\mathbb R_{\ge 0}$, or $\mathbb R_{>0}$ can be treated the same way using the representations:

$$\mathbb R =\bigcup_{i=1}^{\infty} [-n, n], \mathbb R_{\ge 0} =\bigcup_{i=1}^{\infty} [0, n], \mathbb R_{>0} =\bigcup_{i=1}^{\infty} (0, n]. $$

Another proof of Claim 6.

Another proof for Claim 6 can be obtained based on the following result (Corollary 15.2 in Kechris, A. (1995), Classical descriptive set theory. Springer.)

Let $\phi : X \to Y$ be Borel function where $X$ and $Y$ are Borel spaces associated with some Polish spaces (such Borel spaces are called standard Borel spaces). Then, for any Borel set $A \subseteq X$ over which $\phi: A \to Y$ is injective, $\phi(A)$ is Borel in $Y$.

In our problem, the real function $f:D \to \mathbb R$ with $D$ being an interval (bounded or unbounded) is Borel as it is increasing, and injective as it is strictly increasing (Claim 2), so $f(D)$ is Borel by the above result. This proof was suggested by @NateEldredge in a comment here.

Image of an arbitrary cdf

For any discrete random variable, the set of reals $\mathbb R$ can be partitioned into a countably many intervals over which the function is either strictly increasing or stair-step function. Indeed, the union of the images of the function over these intervals is a Borel set. The same can be done for an arbiataray cdf. From the Lebesgue's decomposition theorem's the cdf $F_X$ of any random variable $X$ can be written as

$$F_X(x)=\alpha F_{C}(x)+(1-\alpha ) F_D(x)$$

for some $\alpha \in [0,1]$ where $F_{C}(x)$ and $F_D(x)$ are cdfs of some continuous and discrete distributions, respectively. Based on the discrete part $F_D(x)$, the set of reals $\mathbb R$ can be partitioned into a countably many intervals over which $F_D(x)$ is either strictly increasing or stair step. Hence, $F_X(x)$ on each of these intervals is either strictly increasing or a stair-like function (in which each stair is a continuous function), and similarly we conclude that the image of $F_X(x)$ cdf is Borel.