Left-continuous function that agrees with BV function almost everywhere 
Let $f$ be a a bounded variation (BV) function on the whole real line. Identify a left-continuous function $g$ that agrees with $g$ at all but countably many points, and give an example showing that $g$ may have smaller total variation than $f$.

I know that $f$ can be written as a difference of two non-decreasing function $f_1$ and $f_2$. Every non-decreasing function has only countable discontinuities, so $f_1$ and $f_2$ has only countable discontinuities, and so too does $f$.
For any function $f$ with countable discontinuities, is it possible to find a left-continuous function (i.e. $f(a^-)=f(a)$) that agrees with it at all but countably many points? 
 A: Since $f$ is of bounded variation, we can write $f=f_1-f_2$, where the $f_k$ are non-decreasing.
Since the $f_k$ are non decreasing, the limit $g_k(x) = \lim_{\alpha \uparrow x} f_k(\alpha)$ exists everywhere. Note that if $x<y$ then we must have $f_k(x) \le g_k(y)$. Also, we must have $g_k(x) \le f_k(x)$.
The $g_k$ are non-decreasing (since the $f_k$ are) and continuous from the left.
To see left continuity, choose $x$ and let $\epsilon>0$. By definition of $g_k$, for some $\delta>0$, we have $f_k(y) > g_k(x) -\epsilon$ for all $y \in [x-\delta,x]$. Then if $y \in (x-\delta,x]$ (note the interval is half-open now), we have $g_k(y) \ge f_k(x-\delta) > g_k(x) -\epsilon$. Since $g_k$ is non-decreasing, we have $g_k(x) = \lim_{\alpha \uparrow x} g_k(\alpha)$.
The $f_k$ have an at most countable set of discontinuities, let these be $\Omega_k$. Then $g_k(x) = f_k(x)$ for all $ x \notin \Omega_k$.
Let $g = g_1-g_2$. Then $g(x)=f(x)$ for all $x \notin \Omega = \Omega_1 \cap \Omega_2$, and $\Omega$ is at most countable. The function $g$ is left continuous since the $g_k$ are.
For an example with smaller variation, let $f=1_{\{0\}}$. Then the above construction gives $g=0$. Then $\operatorname{Var} f = 2$, $\operatorname{Var} g = 0$.
