I need help proving a result regarding multipliers for the HK integral A function $g:[a,b]\rightarrow \mathbb{R}$ is said to be of bounded variation on $[a,b]$ if 
$$
\sup\left\{\sum_{i=1}^n|g(x_i)-g(x_{i-1})|:a = x_0 <\ldots<x_n = b\right\}<+\infty.
$$
If $g:[a,b]\rightarrow\mathbb{R}$ is of bounded variation on $[a,b]$ and $f:[a,b]\rightarrow \mathbb{R}$ is HK integrable on $[a,b]$ then $fg$ is HK integrable on $[a,b]$. I have found proof of this that I am content with.
I need help in proving or disproving the following statement:

Let $g:[a,b] \rightarrow \mathbb{R}$ be a function. Then $fg$ is HK integrable on $[a,b]$ for all HK integrable functions $f:[a,b]\rightarrow \mathbb{R}$ if and only if $\exists
g_0:[a,b]\rightarrow\mathbb{R}$ such that $g_0$ is of bounded variation on $[a,b]$ and $g_0 = g$ everywhere except for a set of Lebesgue measure $0$. 

If my claim is wrong, please let me know. 
 A: Your claim is wrong.
It is true that if $g_0$ is of bounded variation on $[a,b]$ and $g = g_0$ up to a set of Lebesgue measure $0$, then $fg$ is HK integrable for any HK integrable $f$ (and I have the feeling you have proven this direction). However, the proof of that direction also proves the following stronger statement. Let $g$ be such that for all $\epsilon > 0$, there are $\{(a_n,b_n) : n \ge 1\}$ such that $\sum_n b_n-a_n < \epsilon$ and $g$ is of bounded variation on $[a,b]\setminus\cup_{n \ge 1} (a_n,b_n)$; then $fg$ is HK integrable for any HK integrable $f$. 
Therefore, to disprove the hard direction of your claim, it suffices to exhibit a function $g$ as above such that there is no $g_0$ of bounded variation that is a.e. equal to $g$. An example of such a $g$ is $1_{0}(x)+\sum_{n \ge 1} (1_{[\frac{1}{2n+1},\frac{1}{2n}]}(x)-1_{[\frac{1}{2n+2},\frac{1}{2n+1}]}(x))$, defined on $[0,1]$. Indeed, $g$ is of bounded variation on $[\epsilon,1]$ for any $\epsilon > 0$, while there is no $g_0$ of bounded variation such that $g=g_0$ a.e..
An analogy to all of this is that a function $f$ is Riemann integrable if and only if the set of its discontinuities has Lebesgue measure $0$. This is very different from $f$ being almost everywhere equal to a continuous function. For example, $f = -1$ for $-1 < x < 0$ and $f = 1$ for $0 \le x < 1$ is of course Riemann integrable, but is not a.e. equal to a continuous function.
Your claim should be:

Let $g:[a,b] \rightarrow \mathbb{R}$ be a function. Then $fg$ is HK integrable on $[a,b]$ for all HK integrable functions $f:[a,b]\rightarrow \mathbb{R}$ if and only if for all $\epsilon > 0$, there are $\{(a_n,b_n) : n \ge 1\}$ such that $\sum_n b_n-a_n < \epsilon$ and $g$ is of bounded variation on $[a,b]\setminus\cup_{n \ge 1} (a_n,b_n)$.

However, this is a different question, so feel free to ask it in another post. (I'll think about it).
