Suppose $f,g$ are two absolutely continous functions in an interval $[a,b]$. Show $fg \in AC[a,b]$. 
Suppose $f,g$ are two absolutely continous functions in an interval $[a,b]$. Show $fg \in AC[a,b]$.

My try:
$$
\begin{align}
\int_a^x (fg)'(t)dt & \stackrel{(1)}= \int_a^x \Big[f'(t)g(t) + f(t)g'(t)\Big]dt \\[8pt]
& = \int_a^x f'(t)g(t)dt  + \int_a^xf(t)g'(t)dt \\[8pt]
& \stackrel{(2)}= f(x)g(x) - f(a)g(a) - \int_a^xf(t)g'(t)dt + \int_a^xf(t)g'(t)dt \\[8pt]
& = f(x)g(x) - f(a)g(a)
\end{align}
$$
This should conclude the proof but I'm not very sure if I can use the product rule in $(1)$ and if I can integrate by parts the first integral in $(2)$.
Is there any other way to prove this fact?
 A: We can prove directly.
Choose $M>0$ such that $|f(x)|\leq M$ and $|g(x)|\leq M$ for all
$x\in[a,b]$. Let $\varepsilon>0$ be given. Since $f$ and $g$ are
absolutely continuous, there exists $\delta>0$ such that for any
(fintely many) pairwisely disjoint subintervals $(x_{1},y_{1}),\ldots,(x_{n},y_{n})$
of $[a,b]$, if $\sum_{i=1}^{n}(y_{i}-x_{i})<\delta$, then $\sum_{i=1}^{n}|f(y_{i})-f(x_{i})|<\frac{\varepsilon}{2M}$
and $\sum_{i=1}^{n}|g(y_{i})-g(x_{i})|<\frac{\varepsilon}{2M}$. Now,
let $(x_{1},y_{1}),\ldots,(x_{n},y_{n})$ be pairwisely disjoint subintervals
of $[a,b]$ with $\sum_{i=1}^{n}(y_{i}-x_{i})<\delta$. For each $i$,
observe that
\begin{eqnarray*}
 &  & \left|f(y_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
 & \leq & \left|f(y_{i})g(y_{i})-f(x_{i})g(y{}_{i})\right|+\left|f(x_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
 & \leq & M\left|f(y_{i})-f(x_{i})\right|+M\left|g(y_{i})-g(x_{i})\right|.
\end{eqnarray*}
It follows that
\begin{eqnarray*}
 &  & \sum_{i=1}^{n}\left|f(y_{i})g(y_{i})-f(x_{i})g(x_{i})\right|\\
 & \leq & M\sum_{i=1}^{n}\left|f(y_{i})-f(x_{i})\right|+M\sum_{i=1}^{n}\left|g(y_{i})-g(x_{i})\right|\\
 & < & \varepsilon.
\end{eqnarray*}
Therefore $fg$ is absolutely continuous.
A: We have $$fg={1\over 4}[(f+g)^2-(f-g)^2]$$ The sum and the difference of AC functions is an AC function. Therefore it suffices to prove that if $h$ is an AC function so is $h^2.$ The latter follows immediately from the inequality $$|h(x_2)^2-h(x_1)^2|\\ =|h(x_2)+h(x_1)|\,|h(x_2)-h(x_1)|\\ \le 2M|h(x_2)-h(x_1)|$$ where $M=\displaystyle\max_{a\le t\le b}|h(t)|.$
A: Let $f,g$ be AC on $I = [a,b]$. By EVT we know that $|f(x)|\le C_f$ and $|g(x)|\le C_g$ then $|f(x)| \& |g(x)| \le \max(C_f, C_g) = C$ where $C_f,C_g, C$ are constants.
Then we have that to every $\epsilon>0$ there corresponds a $\delta>0$ so that
\begin{equation*}
    \begin{aligned}
    \sum_{i=1}^{n} |f(\beta_i) - f(\alpha_i)| < \frac{\epsilon}{2C} \quad \sum_{i=1}^{n} |g(\beta_i) - g(\alpha_i)| < \frac{\epsilon}{2C} 
    \end{aligned}
\end{equation*}
for any $n$ and any disjoint collection of segments $(\alpha_1, \beta_1), \dots, (\alpha_n, \beta_n)$ in $I$
whose lengths satisfy
\begin{equation*}
    \begin{aligned}
    \sum_{i=1}^{n} (\beta_i - \alpha_i) < \delta
    \end{aligned}
\end{equation*}
Then for $h(x)=f(x)g(x)$ on $I$ we have that
\begin{equation*}
    \begin{aligned}
        \sum_{i=1}^{n} |h(\beta_i) - h(\alpha_i)|  &= \sum_{i=1}^{n} \left| f(\beta_i)g(\beta_i) - f(\alpha_i)g(\alpha_i) \right| \\
        &= \sum_{i=1}^{n} \left| f(\beta_i)g(\beta_i) - f(\beta_i)g(\alpha_i) + f(\beta_i)g(\alpha_i) - f(\alpha_i)g(\alpha_i) \right| \\
        &\le \sum_{i=1}^{n} \left| f(\beta_i)g(\beta_i) - f(\beta_i)g(\alpha_i) \right| + \sum_{i=1}^{n} \left| f(\beta_i)g(\alpha_i) - f(\alpha_i)g(\alpha_i) \right| \\
        &= \sum_{i=1}^{n} \left| f(\beta_i)(g(\beta_i) - g(\alpha_i)) \right| + \sum_{i=1}^{n} \left| g(\alpha_i)(f(\beta_i) - f(\alpha_i)) \right| \\
        &\le C \sum_{i=1}^{n} \left| g(\beta_i) - g(\alpha_i) \right| + C \sum_{i=1}^{n} \left| f(\beta_i) - f(\alpha_i) \right| \\
        &\le C \frac{\epsilon}{2C} + C \frac{\epsilon}{2C} = \epsilon \\
    \end{aligned}
\end{equation*}
For every $\epsilon>0$ there corresponds a $\delta$ such that the above condition on $\delta$ holds.
Hence $h = fg$ is AC on $I$.
