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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?

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  • $\begingroup$ (1) holds because both $f$ and $g$ are differentiable (a.e.). As long as both $f'(t)$ and $g'(t)$ exist, we always have $(fg)'(t) = f'(t)g(t) + f(t)g'(t)$. $\endgroup$ – Danny Pak-Keung Chan Jan 28 at 18:18
  • $\begingroup$ However, integration-by-part formula requires a proof. $\endgroup$ – Danny Pak-Keung Chan Jan 28 at 18:19
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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.

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