# Is the following function absolutely continuous?

Following problem is from Exercise $14.15$ of F. Bass' Real Analysis.

Let $(\Omega,\Sigma,\mu)$ be a measure space with $\mu(\Omega)<\infty$ and $f$ be a real-valued integrable function. Define $$g(x)=\int_\Omega|f(y)-x|~d\mu(y)$$ for $x\in\mathbb{R}$. Show that $g$ is absolutely continuous. Find $g^\prime(x)$ and prove that $g(x_0)=\inf_{x\in\mathbb{R}}g(x)$ if and only if $$\mu(\{y:f(y)>x_0\})=\mu(\{y:f(y)<x_0\}).$$

According to the definition of absolutely continuous functions, I can prove $g$ is absolutely continuous. But I'm puzzled by the second question about derivatives. Any help is appreciated!

• If $\mu ( \Omega) = \infty,$ then $g(x) = \infty$ for all $x\ne 0.$ – zhw. Oct 26 '17 at 4:43
• Alright then. So the original statement should assume that $\Omega$ has finite measure. – Jay Oct 26 '17 at 4:52

## 1 Answer

Hint:

Proceeding formally you would guess that $g'(x) = -\int_\Omega \operatorname{sgn}(f(y)-x)\,d\mu(y)$. Can you prove that by integrating this guess you recover $g$? (Try Fubini.) If so then it confirms both the absolute continuity and the formula for $g'$.

For the part about the minimum, it might help to observe that if the above guess is right, then the last statement holds for $x_0$ iff $g'(x_0) = 0$. And you would expect the derivative of $g$ to vanish at the minimum of $g$.