Let $f$ be absolutely continuous and show the function $g(x)=\int_{0}^{1}f(xy)dy$ is absolutely continuous Let $f$ be absolutely continuous and show the function $$g(x)=\int_{0}^{1}f(xy)dy$$ is absolutely continuous.
I know the proper solution which simply uses the definition of absolute continuity. However, I cannot see what is wrong with this approach, that shows the function is not even continuous.
We simply use $u$ substitution to get $g(x)=\frac{1}{x}\int_{0}^{1}f(u)du$ except $g(0)=0$. This to me is clearly not continuous, what is wrong here?
 A: When you used the substitution $u = xy$, you have $du = xdy \implies dy = \frac{du}{x}$. This is all fine, but you didn't adjust the upper integral limit. In particular, at $y = 1$, you have $u = x$ so the proper result would be, for $x \neq 0$,
$$g(x) = \frac{1}{x}\int_{0}^{x}f(u)du \tag{1}\label{eq1A}$$
and with $g(0) = 0$. I trust you can see that $g(x)$ can now potentially be continuous with this corrected definition.
A: Let $(a_i,b_i)$ for $i=1,\ldots,n$ be a set of disjoint intervals of $[0,1]$. Then $\sum_{i=1}^{n}|g(b_i)-g(a_i)|=\sum_{i=1}^{n}|\int_{0}^{1}f(b_iy)-f(a_iy)dy|\leq\sum_{i=1}^{n}\int_{0}^{1}|f(b_iy)-f(a_iy)|dy=\int_{0}^{1}\sum_{i=1}^{n}|f(b_iy)-f(a_iy)|dy$
Then, for any $\varepsilon>0$, there exists $\delta>0$ such that $\sum_{i=1}^{n}(b_i-a_i)<\delta$, since $y\in(0,1)$, $\sum_{i=1}^{n}(b_iy-a_iy)<\sum_{i=1}^{n}(b_i-a_i)<\delta$ implies $\sum_{i=1}^{n}|f(b_iy)-f(a_iy)|<\varepsilon$ by the absolute continuity of $f$. Hence, $\sum_{i=1}^{n}|g(b_i)-g(a_i)|<\int_{0}^{1}\varepsilon dy=\varepsilon$ which is exactly the definition of $g$ being absolutely continuous.
