real analysis - Intermediate Value Property + BV nearly everywhere = continuous?

It is well known that a function $f:[a,b]\to\mathbb R$ which has the intermediate value property (i.e. if $a',b'\in[a,b]$, then for each $c$ between $f(a')$ and $f(b')$ there is some $x$ between $a'$ and $b'$ such that $f(x)=c$) cannot have jump discontinuities, and that a function $g:[a,b]\to\mathbb R$ of bounded variation can only have jump discontinuities. Therefore, a function having the intermediate value property and being of bounded variation must be continuous. I wonder if the BV-condition can be relaxed a little, so my question is:

Is there a discontinuous function $f:[a,b]\to\mathbb R$ and a function $g:[a,b]\to\mathbb R$ of bounded variation, such that $f$ has the intermediate value property and agrees with $g$ at every point of $[a,b]$ except at countably many?

Any hints/help is highly appreciated. Thanks in advance!

Answer

Let $x\in[a,b]$ be a discontinuity of $f$.

We assume without loss of generality that $x\in(a,b)$, the case of the boundary can be dealt with similarly.

Since $g$ is of bounded variation, the limits $l=\lim_{y\to x^-}g(y)$ and $L=\lim_{y\to x^+}g(y)$ exist and are finite.
On the other hand, since $f$ is Darboux and discontinuous at $x$, at least one of $\lim_{y\to x^-}f(y)$ and $\lim_{y\to x^+}f(y)$ must not exist, say the latter.

We hence have that

\begin{align}\forall \epsilon>0,\,&\exists\delta>0,\,\forall y \in(x,x+\delta),\,\,|g(y)-L|<\epsilon\tag{1}\\
\exists \epsilon_0>0,\,&\forall\delta>0,\,\exists y \in(x,x+\delta),\,\,|f(y)-L|\geq\epsilon_0\tag{2}\end{align}

Take $\epsilon=\frac{\epsilon_0}{2}$ in $(1)$, so that there is some $\delta_0>0$ with the property that $|g(y)-L|<\frac{\epsilon_0}{2}$ whenever $y \in (x,x+\delta_0)$.
By $(2)$, there is some $y_1\in (x,x+\delta_0)$ with $|f(y_1)-L|\geq \epsilon_0$.

If $g=f$ except on a countable subset of $(x,x+\epsilon)$, there must be some $y_2\in(x,y_1)$ with $g(y_2)=f(y_2)$.
Hence, $|g(y_2)-L|=|f(y_2)-L|<\frac{\epsilon_0}{2}$.

Now, since $f$ is Darboux, it must assume all values between $f(y_1)$ and $f(y_2)$ on $(y_1,y_2)$.
In particular, it must assume an uncountable range of values $v$ with $\frac{\epsilon_0}{2}\leq |v-L|\leq \epsilon_0$.
This means there is some $y_3\in(y_2,y_1)\subset(x,x+\delta_0)$ with $\frac{\epsilon_0}{2}\leq |g(y_3)-L|\leq \epsilon_0$, which contradicts the defining property of $\epsilon_0$. $\square$