Does $0$ belong to the Lebesgue set of Heaviside function?

157 Views Asked by At

Lebesgue set: For a function $f \in L^1_{\operatorname{loc}} (R^n)$ and $x \in R^n$, $x$ is said to be a point in the Lebesgue set of $f$ if there exists a number $A$ such that

$\lim_{r=0} 1/(\lambda(B(x,r)) \int_{B(x,r)} |f(y)-A|dy = 0$.

Now if we consider $H$ to be the heaviside function

$H(x) = \begin{cases} 1 & \text{if } x > 0 , \\ 1/2 & \text{if } x = 0, \\ 0 & \text{if } x<0. \end{cases}$

I have been able to show that $H(x) = \lim_{r=0} 1/2r \int_{x-r}^{x-r} H(y)dy$

But from this, it follows that $H(0) = 1/2$, and thus $0$ belongs in the lebesgue set. What am I doing wrong here?

1

There are 1 best solutions below

3
On

No matter what $A$ is, one of $|0-A|\ge\frac12$ or $|1-A|\ge\frac12$ will hold. Therefore, $$\int_{-r}^r|H(x)-A|\,dx\ge\frac r2, \quad\text{and so}\quad \frac1{2r}\int_{-r}^r|H(x)-A|\,dx\ge \frac14. $$