How to prove that this function is continuous and monotone?

Question

I am studying Measure Theory using the book by Elstrodt and there are a few statements in the text in the section on contents on the semi-ring $\mathfrak{J} := \{\ ]a,b]\ :\ a \leq b,\ a,b \in \mathbb{R}\ \}$ for which I'm having trouble finding a proof.

The the book states:

Every increasing, continuous function defines a finite pre-measure $\mu_F: \mathfrak{I} \to \mathbb{R}$. Such a pre-measure can be seen as a continuous mass distribution on $\mathbb{R}$. However, there are also completely different pre-measures:

Let $A \subset \mathbb{R}$ be a countable set and $p: A \to \mathbb{R}$ a strictly positive function such that

$$\sum \limits_{y \in A \cap [-n,n]} p(y)$$

converges for all $n \in \mathbb{N}$.

Then

$$\mu(]a,b]) := \sum \limits_{y \in A \cap ]a,b]} p(y) \quad (a \leq b)$$

is a finite pre-measure on $\mathfrak{I}$, and so

$$G(x) := \begin{cases} \sum \limits_{y \in A \cap ]0,x]} p(y), & \text{for } x \geq 0\\ -\sum \limits_{y \in A \cap ]x,0]} p(y), & \text{for } x < 0 \end{cases}$$

is right-continuous. We call such a function a jump function.

The function $G$ is discontinuous exactly at $x \in A$ (see my previous question).

Now here comes the part I am struggling with:

Now consider any increasing, right-continuous function $F: \to \mathbb{R}$ and let $A$ be the set of discontinuities of $F$. Then $A$ is countable since $A = \bigcup \limits_{n=1}^{\infty} A_n$ where $A_n = \{x \in [-n,n]: \lim \limits_{h \downarrow 0} (F(x+h) - F(x-h)) \geq \frac{1}{n}\}$ are finite due to the monotonicity of $F$. For $y \in A$, let $p(y) = \lim \limits_{h \downarrow 0} (F(x+h) - F(x-h))$. Now if $y_1,...,y_n \in A \cap ]-n,n[$ are distinct, then $\sum \limits_{j=1}^{k} p(y_j) \leq F(n) - F(-n)$. Hence, $\sum \limits_{y \in A \cap [-n,n]} p(y)$ converges for all $n \in \mathbb{N}$. Now let $G$ be the corresponding jump function, then $H = F - G$ is continuous on $\mathbb{R}$ and increasing. This means $F$ can be expressed as $F = G + H$, i.e. it is the sum of an increasing jump function and a continuous, increasing function. The functions $G$ and $H$ are unique up to additive constants.

Now most of the last paragraph is clear to me (see here for a proof that $A_n$ is finite and for the convergence of the series of $p(y)$). But how can I prove that

1. $H$ is continuous,

2. $H$ is increasing and

3. that the decomposition of $F$ into $G$ and $H$ is unqiue up to additive constants?

Proof attempt:

1) Continuity of $H$:

Obviously, $H$ is continuous at $x \notin A$ as the difference of two continuous functions and inuitively it should also be continuous at $x \in A$ since $G$ is constructed to remove the discontinuities of $F$ ($p(y) = \lim \limits_{h \downarrow 0} (F(x) - F(x-h)$ since $F$ is right-continuous).

To show that $H = F - G$ is left-continuous at $y \in A$ I have to show that

$$\lim \limits_{n \to \infty} (F(y) - F(x_n) - (G(y) - G(x_n))) = 0$$

for all sequences $x_n \to y$. If $x_n = y$ for any $n \in \mathbb{N}$, then this clearly holds. So it remains to show that $\lim \limits_{n \to \infty} G(y) - G(x_n) = p(y)$ for all sequences $x_n \to y$ with $x_n < y$ for all $n \in \mathbb{N}$.

From my previous question I know that $\lim \limits_{n \to \infty} G(y) - G(x_n) \geq p(y)$ for all sequences $x_n \to y$ with $x_n < y$ for all $n \in \mathbb{N}$, but how can I prove that this must be an equality?

2) $H$ is increasing:

To prove that $H$ is increasing I need to show that

$$F(y) - F(x) - (G(y) - G(x)) \geq 0$$

if $y > x$ (the case $y=x$ is trivial). In other words I need to show that $F$ grows faster than $G$.

Sorry for the long post and thanks a lot!

Answer 1

1. Continuity

Proving continuity is much easier than you think. From your previous questions, I can see that you already know $G$ is right continuous, but you seem confused on why we must have $\lim_{h\downarrow 0}G(x)-G(x-h)=p(x)$. These two facts can actually be proven in the same exact way. Firstly, for $x<y$, I assume you'll accept the following equality. $$G(y)-G(x) = \sum_{t\in A\cap]x,y]}p(t)$$

This form is hard to use, since the domain of summation depends on $x$ and $y$. To fix this, we'll use an indicator function. Let $I(x,t,y)=1$ whenever $t\in ]x,y]$, and $I(x,t,y)=0$ otherwise, then we have the following equalities. $$\begin{align} G(y)-G(x) &= \sum_{t\in A\cap]x,y]} p(t)\cdot I(x,t,y) \\ &= \sum_{t\in A} p(t)\cdot I(x,t,y) \end{align}$$

The first equality holds since the summand is unchanged on that domain, and the second equality holds since extending the domain now just adds a bunch of zeros. This form is extremely convenient however, since it lets us resolve directional limits. For convenience, we extend the domain of $p$ so that $p(t)=0$ whenever $t\notin A$, then the following holds. $$\begin{align} \lim_{x\uparrow y}G(y)-G(x) &= \lim_{x\uparrow y} \sum_{t\in A} p(t)\cdot I(x,t,y) \\ &= \sum_{t\in A} \lim_{x\uparrow y} p(t)\cdot I(x,t,y) \\ &= p(y) \end{align}$$

Swapping the limit with the summation follows directly from the Dominated Convergence theorem. The summand is bounded by the absolutely summable function $p\cdot I(y-1,t,y)$, and moreover the summand converges pointwise. Namely, the summand converges to $0$ when $t\neq y$, and converges to $p(y)$ when $t=y$, hence the sum of limits is just $p(y)$. Incidentally, this discrete version of Dominated Convergence is very easy to prove. The above technique can also be modified to give a simpler proof that $G$ is right-continuous. Since $G$ is right-continuous by proof, and $F$ is right-continuous by premise, then $H=F-G$ is also right-continuous. To prove $H$ is left-continuous, we just resolve the following limit.

$$\begin{align} \lim_{h\downarrow 0} H(x+h)-H(x-h) &= \left(\lim_{h\downarrow 0}F(x+h)-F(x-h)\right) - \left(\lim_{h\downarrow 0}G(x+h)-G(x-h)\right) \\ &= p(x)-p(x) = 0 \\ \lim_{h\downarrow 0} H(x-h) &= \lim_{h\downarrow 0}H(x+h) = H(x) \end{align}$$

So $H$ is also left-continuous, therefore $H$ is continuous.

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2. Monotonicity

Proving $H$ is monotonic actually follows from the same argument which was used to show $p$ is summable. Given values $x<a_1<a_2<\cdots<a_N\leq y$, we easily prove $\sum_k p(k) \leq F(y)-F(x)$. Since the difference $G(y)-G(x)$ is the supremum of all those finite sums, we get $G(y)-G(x)\leq F(y)-F(x)$ leading to $H(x)\leq H(y)$.

To carry out this argument more formally, take any finite set $S\subseteq ]x,y]$, find a strictly increasing enumeration $a_1<a_2<\cdots<a_N$ of the set $S=\{a_1,\cdots,a_N\}$, and then observe the following. $$\begin{align} \sum_{t\in S} p(t) &= \sum_{n=1}^N p(a_n) \\ &= \sum_{n=1}^N \lim_{h\downarrow 0} F(a_n)-F(a_n-h) \\ &= \lim_{h\downarrow 0} \left(F(a_N)+\sum_{n=1}^{N-1} F(a_n)\right)-\left(F(a_1-h)+\sum_{n=2}^N F(a_n-h)\right) \\ &\leq \lim_{h\downarrow 0} \left(F(y)+\sum_{n=1}^{N-1} F(a_n)\right)-\left(F(x)+\sum_{n=2}^N F(a_{n-1})\right) \\ &= F(y)-F(x) \end{align}$$

The above inequality holds for all sufficiently small $h>0$, since we have $x<a_1<a_2<\cdots<a_N\leq y$ and since $F$ is monotonic. The above now leads to the following, proving that $H$ is monotonic increasing. $$G(y)-G(x) = \sup_{S\subseteq ]x,y]}^{\#S<\infty} \sum_{t\in S}p(t) \leq F(y)-F(x)$$ $$H(y)-H(x) = (F(y)-F(x))-(G(y)-G(x)) \geq 0$$

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3. Uniqueness

Finally, we show that the decomposition $F=G+H$ is unique. Let $G'$ be any increasing jump function, such that the difference $H':=F-G'$ is continuous. Since $G'$ is an increasing jump function, there's a set $A'$ and function $p':A'\to\mathbb{R}$ such that the following holds for all $x<y$. $$G'(y)-G'(x)=\sum_{t\in A'\cap]x,y]} p'(t)$$

For convenience we let $U=A\cup A'$, which is countable, and extend $p'$ so that $p'(t)=0$ whenever $t\notin A'$. We now easily get the following equalities. $$\begin{align} G'(y)-G'(x) = \sum_{t\in U\cap]x,y]} p'(t) \\ G(y)-G(x) = \sum_{t\in U\cap]x,y]} p(t) \\ \end{align}$$

To prove uniqueness, we simply prove $p'=p$. This is trivial, since the continuity of $H$ and $H'$ leads immediately to the following. $$\begin{align} p(x) &= \lim_{h\downarrow 0}(G(x+h)-G(x-h)) \\ &= \lim_{h\downarrow 0}(F(x+h)-F(x-h)) \\ &= \lim_{h\downarrow 0}(G'(x+h)-G'(x-h)) \\ &= p'(x) \end{align}$$

We see that $p=p'$, and consequently $G'(y)-G'(x) = G(y)-G(x)$, from which we infer $G-G'$ is constant. It follows that $H+G=F = H'+G'$ and consequently $H'-H=G-G'$ is the exact same constant. This proves the decomposition $F=G+H$ unique, up to an additive constant.

Note: Technically, the provided definition of jump functions seems to require that $G(0)=0$, in which case we actually get $G=G'$ and $H=H'$. I suspect this is unintended, a minor error induced by the quoted text being overly vague about the exact definition of "jump function".