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An function $f$ is absolutely continuous if for each $\varepsilon$ there is $\delta$ such that if $(a_i,b_i)_i$ are disjoint open intervals such that $\sum|a_i-b_i|<\delta$ then $\sum|f(a_i)-f(b_i)|< \varepsilon$.

I know that for two absolutely continuous functions on compact $X$, $f,g: X \to \mathbb{R}$, their product $fg$ is also absolutely continuous. However, the proof relies on $f$ and $g$ being bounded - not necessarily so if we have $f,g: \mathbb{R}\to\mathbb{R}$. Is there a counterexample on this domain where $fg$ is not absolutely continuous?

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Let $f(x)=g(x)=x$. While $f$ and $g$ are absolutely continuous on $\mathbb{R}$, their product $f(x)g(x)=x^2$ is not, because it is not uniformly continuous on $\mathbb{R}$.

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I think that $f(x)=x=g(x)$ is a counterexample on $\mathbb R$.

It is trivial to argue that $\delta:=\epsilon$ shows that each is absolutely continuous.

Now, let us argue that $f$ is not absolutely continuous. Assume by contradiction it is.

Let $\epsilon >0$ (you could pick $\epsilon=1$ if you want). Then, there exists $\delta$ with the given property. Pick the interval $I=(n, n+\delta/2)$, and hence $$|f(n+\frac{\delta}{2})-f(n)| <\epsilon \Rightarrow \\ n\delta+ \delta^2/4 < \epsilon \Rightarrow \\ n\delta <\epsilon$$

By picking the right $n$ you geta contradiction.

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