Let $f$ and $g$ be a locally integrable function on $[0,1]$. For any $\phi\in C^1[0,1]$, $f$ and $g$ satisfies
$\displaystyle \int_{0}^{1}\phi'(t)f(t)dt+\int_{0}^{1}\phi(t)g(t)dt = \phi(1)f(1)-\phi(0)f(0) \tag{1}$
I would like to prove that $f$ is absolutely continuous (or prove otherwise).
My Attempt
The above equation basically states that $g$ is the distributional derivative of $f$. So I tried to follow When is the distributional derivative a function? to prove that $f$ is absolutely continuous.
We let $u(t) =\int_{0}^{t}g(t)dt+f(0)$. Since $u$ is an indefinite integral, it is absolutely continuous, and $u'(t)=g(t)$ almost everywhere. Moreover, we have $u(0)=f(0), u(1)=f(1)$. We substitute $\phi(t)=1$ to Equation (1) to see that $\int_{0}^{1}g(t)dt = f(1)-f(0)$.
Now, by integration by parts,
$\begin{align} \int_{0}^{1}\phi(t)g(t)dt &= \int_{0}^{1}\phi(t)u'(t)dt \\ &= \phi(1)u(1)-\phi(0)u(0)-\int_{0}^{1}\phi'(t)u(t)dt\end{align}$
Substitute this to Equation (1) to see that
$\displaystyle \int_{0}^{1}\phi'(t)f(t)dt+\phi(1)u(1)-\phi(0)u(0)-\int_{0}^{1}\phi'(t)u(t)dt = \phi(1)f(1)-\phi(0)f(0)$
Therefore
$\displaystyle \int_{0}^{1}\phi'(t)f(t)dt = \int_{0}^{1}\phi'(t)u(t)dt$
Since this is true for any $\phi\in C^1[0,1]$, we have $f(t)=u(t)$ almost everywhere. ...(A)
End Attempt
I have two questions regarding the above.
(1) Regarding (A), that is, $\int \phi' fdt = \int \phi' udt \Rightarrow f=u \textrm{ a.e.}$, I think this should be true, but I don't know the proof. Is this even true, and if it is true, how do I prove that?
(2) $f(t)=u(t)$ almost everywhere does not imply that $f$ is absolutely continuous, right? Can we say anything nice about $f$ e.g., $f$ is differentiable a.e.?
If $\psi \in C[0,1]$ we can take $\phi (x)=\int_0^{x} \psi (t)dt$ to see that $\int \psi (f-u)=0$. Let $0<a<1$ and $\psi_n (x)=1$ for $x <a$, $0$ for $x >a+\frac 1 n$ and let $\psi$ have a straight line graph between $a$ and $a+\frac 1n$. Taking limits in $\int \psi_n (f-u)=0$ conclude that $\int_0^{a} (f-u)(t)dt=0$. Thus, $\int_a^{b} (f-u)(t)dt=0$ whenever $a<b$. This implies $f=u$ a.e.
Clearly, $f=u$ a.e. does not even imply continuity of $f$. What we can say is $f$ is a.e. equal to a function which is absolutely continuous. Elements of $L^{1}$ are equivalence classes and $f$ is same as the absolutely continuous function $u$ as an element of $L^{1}$.