Let ${F: {\mathbb R} \rightarrow {\mathbb R}}$ be of bounded variation. Define the positive variation ${F^+: {\mathbb R} \rightarrow {\mathbb R}}$ of ${F}$ by the formula

$\displaystyle F^+(x) := \sup_{x_0 < \ldots < x_n \leq x} \sum_{i=1}^n \max(F(x_{i}) - F(x_{i-1}),0)$

and the negative variation $F^-$ by

$\displaystyle F^-(x) := \sup_{x_0 < \ldots < x_n \leq x} \sum_{i=1}^n \max(F(x_{i-1}) - F(x_{i}),0)$

Establish the identity

$\displaystyle F(x) = F(-\infty) + F^+(x) - F^-(x)$, where ${F(-\infty) := \lim_{x \rightarrow -\infty} F(x)}$.(Hint: The main difficulty comes from the fact that a partition ${x_0 < \ldots < x_n \leq x}$ that is good for ${F^+}$ need not be good for ${F^-}$, and vice versa. However, this can be fixed by taking a good partition for ${F^+}$ and a good partition for ${F^-}$ and combining them together into a common refinement.)

From the hint,I assume we want good partitions $x_0 < \ldots < x_n \leq x$ for $F^-$ and $y_0 < \ldots < y_m \leq x$ for $F^+$ respectively, such that $F(x) < F(- \infty) + F^+(x) - \sum_{i=1}^n \max(F(x_{i-1}) - F(x_i), 0) < F(x) + \varepsilon_1$ and $F(x) - \varepsilon_2 < F(- \infty) + \sum_{i=1}^m \max(F(y_{i}) - F(y_{i-1}, 0) - F^-(x) < F(x)$. We then form the common refinement and take the sup so that $\varepsilon_1, \varepsilon_2 \rightarrow 0$. Yet I’m not sure how to explicitly construct these partitions. Am I confusing what the hint is suggesting at?


1 Answer 1


Let's fix $x\in\mathbb{R}$ for the rest of the proof. By analogy with Riemann sums, I'm going to call a partition any finite sequence $X=(x_i)_{i=0,\ldots,n}$ such that $x_0<\ldots<x_n\leq x$. We'll also say that a partition $X'$ is a refinement of $X$ if every element of $X$ is also in $X'$.

For a partition $X=(x_i)_{i=0,\ldots,n}$, let's define the sums: $$ S^+(X) = \sum_{i=1}^n \max\left(F(x_i) - F(x_{i-1}),0\right) ,\ \ \ \ \ \ \ S^-(X) = \sum_{i=1}^n \max\left(F(x_{i-1}) - F(x_i),0\right). $$ Those are just the sums present in definitions of $F^+$ and $F^-$, this will make notation easier. We can see that we have $F^+(x)=\sup_X S^+(X)$ and $F^-(x)=\sup_X S^-(X)$. By definition of the supremum, we also have for any partition $X$, $S^+(X) \leq F^+(x)$ and $S^-(X) \leq F^-(x)$.

Now here's an important property that might not be obvious at first: let $X'$ be a refinement of another partition $X$. Then $S^+(X)\leq S^+(X')$ and $S^-(X)\leq S^-(X')$.

To see that this is true, consider a partition $X=(x_i)_{i=0,\ldots,n}$, and for some $i$, insert an element $x_{i-1/2}$ between $x_{i-1}$ and $x_i$. Then: \begin{align*} \max\left(F(x_i) - F(x_{i-1}),0\right) &= \max\left(F(x_i) - F(x_{i-1/2}) + F(x_{i-1/2}) - F(x_{i-1}),0\right) \\ &\leq \max\left(F(x_i) - F(x_{i-1/2}),0\right) + \max\left(F(x_{i-1/2}) - F(x_{i-1}),0\right). \end{align*} Similarly, inserting an element less than $x_0$ or greater than $x_n$ just adds a non-negative term to the sum, so by recursion we get $S^+(X)\leq S^+(X')$ for any refinement $X'$ of $X$ (we can use the same arguments for $S^-$).

$ $

Now, on to the main proof. Let's fix some $\varepsilon>0$. By definition of the supremum, we know there exist partitions $X^+$ and $X^-$ such that: $$ 0 \leq F^+(x) - S^+(X^+) < \varepsilon, \ \ \ \ \ \ \ \ 0 \leq F^-(x) - S^-(X^-) < \varepsilon. $$ We can also choose some $\bar{x}\in\mathbb{R}$ far enough to the left so that $|F(-\infty)-F(\bar{x})|<\varepsilon$.

Now let's define a new partition $X=(x_i)_{i=0,\ldots,n}$ that's made up of every element of $X^+$, every element of $X^-$, $\bar{x}$, and $x$. $X$ is a refinement of $X^+$ and $X^-$, so based on what we've shown before: $$ S^+(X^+) \leq S^+(X) \leq F^+(x), \ \ \ \ \ \ \ \ S^-(X^-) \leq S^-(X) \leq F^-(x), $$ which gives us: $$ 0 \leq F^+(x) - S^+(X) < \varepsilon, \ \ \ \ \ \ \ \ 0 \leq F^-(x) - S^-(X) < \varepsilon. $$ Since every element of $X$ is less than $x$ and we've included $x$, we have $x_n=x$. Moreover, we can assume to have chosen $\bar{x}$ far enough to the left so that $x_0=\bar{x}$.

Now that we've set everything up, we can write: \begin{align*} |F(x) - F(-\infty) - F^+(x) + F^-(x)| &\leq |F(x)-F(-\infty) - S^+(X) + S^-(X)| + |S^+(X)-F^+(x)| + |S^-(X)-F^-(x)| \\ &\leq |F(x)-F(-\infty) - S^-(X) + S^-(X)| + 2\varepsilon. \end{align*}

We can see that we have: $$ S^+(X) - S^-(X) = \sum_{i=1}^n F(x_i)-F(x_{i-1}) = F(x_n) - F(x_0) = F(x) - F(\bar{x}), $$ and so: $$ |F(x)-F(-\infty) - S^+(X) + S^-(X)| = |F(-\infty)-F(\bar{x})| < \varepsilon, $$ finally giving us: $$ |F(x) - F(-\infty) - F^+(x) + F^-(x)| < 3\varepsilon. $$ This is true for every $\varepsilon>0$ and for every $x\in\mathbb{R}$, and so we have $F(x) = F(-\infty) + F^+(x) - F^-(x)$.

  • $\begingroup$ The fact that a common refinement increases both $S^+$ and $S^-$ is obvious, by simply noting that points added between any interval on which $F$ is monotone increasing or decreasing don't change the value of $S^+$ and $S^-$ on that interval, and points which fall outside of such intervals contribute new segment of rise and fall, which contribute to the total sum. $\endgroup$
    – shark
    Commented Jul 14, 2023 at 1:04

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