Background information:

If $F:\mathbb{R}\rightarrow \mathbb{C}$ and $x\in\mathbb{R}$, we define $$T_{F}(x) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}$$ $T_F$ is called the total variation of $F$. We observe that the sums in the definition of $T_F$ are made bigger if the additional subdivision points $x_j$ are added. Hence, if $a<b$, the definition of $T_F(b)$ is unaffected if we assume that $a$ is always one of the subdivision points. It follows that $$T_F(b) = T_F(a) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N},a=x_0<\ldots<x_n = b\}$$ Thus $T_F$ is an increasing function with values $[0,\infty]$. If $T_F(\infty) = \lim_{x\rightarrow\infty}T_F(x)$ is finite, we say that $F$ is of bounded variation on $\mathbb{R}$, and we denote the space of all such $F$ by $BV$. And we define $NBV$ $N$ for normalized $$NBV = \{F\in BV:F \ \text{is right continuous and} \ F(-\infty) = 0\}$$

Problem 28 - If $F\in NBV$, let $G(x) = |\mu_F|((-\infty,x])$. Prove that $|\mu_F| = \mu_{T_F}$ by showing that $G = T_F$ via the following steps.

a.) From the definition of $T_F$, $T_F\leq G$

b.)$|\mu_F(E)|\leq \mu_{T_F}(E)$ where $E$ is an interval, and hence $E$ is a Borel set.

c.)$|\mu_F|\leq \mu_{T_F}$, and hence $G\leq T_F$

Attempted proof a.): If $x\in\mathbb{R}^n$ then \begin{align*} T_F(x) &= \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}\\ &= \sup\{\sum_{1}^{n}|\mu_F((-\infty,x_j]) - \mu_F((-\infty,x_{j-1}])|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}\\ &= \sup\{\sum_{1}^{n}|\mu_F((x_{j-1},x_j])|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}\\ &\leq \sup\{\sum_{1}^{n}|\mu_F(E_j)|:\{E_j\}_{1}^{n}\in\mathbb{B}_{\mathbb{R}} \ \text{disjoint such that} \ \bigcup_{1}^{n}E_j = \sum_{1}^{n}(-\infty,x_j]\}\\ & = |\mu_F|(E)\\ &= |\mu_F|((-\infty,x])\\ &= G(x) \end{align*}

Attempted proof b.) We have from part a.) $$T_F(x) = \sup\{\sum_{1}^{n}|\mu_F((x_{j-1},x_j])|:n\in\mathbb{N},-\infty<x_0<\ldots<x_n = x\}$$ and we know that the interval $(x_{j-1},x_j]$ is a half-open interval and thus is in the Borel set.

So, let $\epsilon > 0$, then since $F$ is right continuous and $|\mu_F|(E) = |\mu_F|((-\infty,x])$ there exists a $\delta_x > 0$ such that $$\left|\int|\mu_F|(E)dx\right| = \left|\int|\mu_F|((-\infty,x])dx\right| \leq \int \mu_{T_F}(x)dx + \epsilon$$

I am not really sure where to go from here or if this is even correct, I am pretty sure if I can complete part b.) then c.) should not be too hard. I just need some help with proving part b.), any suggestions is greatly appreciated.


For part (b), you can proceed as follows: Let $(a, b]$ be any interval, let $a = x_0 < x_1 < ... < x_n = b$ be any partition, then $|\mu_{F}((a,b])| \leq \sum_{1}^{n} |\mu_{F}((x_{j-1},x_{j}])| = \sum_{1}^{n} |F(x_{j}) - F(x_{j-1})| \leq \sup\{\sum_{1}^{n} |F(x_{j} - F_{j-1}|: n \in \mathbb{N}, a = x_0 <...<x_n = b\} = T_{F}(b) - T_{F}(a) = \mu_{T_{F}}((a,b])$.


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