关于有限变差函数$F$的$V_F(-\infty,b]=\lim_{a\to-\infty}V_F[a,b]$的证明问询

Hey, let's work through this proof about finite variation functions together. First, let's clearly state the proposition we're dealing with:

Proposition
Suppose that $F:\mathbb{R}\to\mathbb{R}$ is of finite variation. If $b\in\mathbb{R}$, then
$$V_F(-\infty,b]=\lim_{a\to-\infty}V_F[a,b].$$

From what you shared, here's the proof (I'll fill in the logical gaps where your original text cut off, since this is a standard result):

Proof
Let $\epsilon > 0$. By the definition of total variation over $(-\infty, b]$, we can pick an increasing sequence ${t_i}{i=0}^n$ of points in $(-\infty, b]$ such that the sum of absolute differences over this sequence is within $\epsilon$ of the total variation:
$$\sum{i=1}^n|F(t_i)-F(t_{i-1})|>V_F(-\infty,b]-\epsilon.$$
Now, take any $a \leq t_0$. Since ${t_i}$ is a partition of $[t_0, b]$, which is a subset of $[a, b]$, this sequence is also a valid partition of $[a, b]$. By the definition of total variation over $[a, b]$, the supremum over all partitions (which is $V_F[a,b]$) must be at least the sum from our chosen partition. So we get:
$$V_F[a,b] \geq \sum_{i=1}^n|F(t_i)-F(t_{i-1})|>V_F(-\infty,b]-\epsilon.$$

Next, we note that for any $a$, $[a, b]$ is a subset of $(-\infty, b]$. This means every partition of $[a, b]$ is also a partition of $(-\infty, b]$, so the total variation over $[a,b]$ can't exceed that over $(-\infty,b]$:
$$V_F[a,b] \leq V_F(-\infty,b].$$

Combining these two inequalities: for any $\epsilon>0$, there exists a $t_0$ such that whenever $a \leq t_0$,
$$V_F(-\infty,b] - \epsilon < V_F[a,b] \leq V_F(-\infty,b].$$
This is exactly the $\epsilon$-definition of the limit $\lim_{a\to-\infty}V_F[a,b] = V_F(-\infty,b]$, so the result holds.

If there was a specific part of the original proof that you found confusing (or if the cut-off section had a different approach), feel free to share more details and we can unpack it further!

备注：内容来源于stack exchange，提问作者Shenron