# 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?

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

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.