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We define the Heaviside step function:

\begin{equation} \mathcal{H}(x) = \mathbb{1} _{[0, \infty[}(x) \end{equation}

Why is $\mathcal{H}$ an element of $L_{loc}^1(\Omega)$? I.e., $\forall K$ compact in $\Omega$, $\mathcal{H} \in L^1(\Omega)$? I've got the following result:

\begin{equation} \int_{\Omega} |\mathcal{H}(x)|dx = \int_{0}^{\infty}dx = \infty \end{equation} which is not strictly inferior to infinity. What am I doing wrong?

  • Any compact set $K$ can be contained in some interval $[-M,M]$ on which the integral of $H$ is $M$. – Ian Oct 10 '16 at 18:51
  • Your definition of $L^1_{loc}$ starts with $\forall K$ but does not use $K$. You have just defined $L^1$. – Ben Oct 10 '16 at 19:02

1 Answers1

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You're writing wrong the condition for locally integrable. It is that $H\in L^1(K)$ for every $K$ compact in $\Omega$, which reduces to prove that $\int_K|H(x)|dx=\int_{K\cap(-\infty,0)}0\cdot dx+\int_{K\cap[0,+\infty)}1\cdot dx=|K\cap[0,+\infty)|\leq|K|<+\infty$

  • Thanks. I have just another question: how do you know that a compact is of finite measure? –  Oct 10 '16 at 19:34
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    It is a theorem that the compact sets on $\mathbb{R}$ are exactly the bounded closed ones. From there it follows that $K\subset [-M,M]$ for some $M>0$ and you're done cause $|[-M,M]|=2M<+\infty$. Another approach would be to use the definition of compact; take the covering ${D_n:=(n-1, n+1):n\in\mathbb{Z}}$. This covers $\mathbb{R}$ and particularly $K$. As $K$ is compact there exists by definition a finite subcovering ${D_{n_1},\ldots,D_{n_k}}$ of $K$, and it follows that $|K|\leq |\cup_j D_{n_j}|\leq\sum_j|D_{n_j}|=2k<+\infty$. –  Oct 10 '16 at 21:38