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Let $(\Omega,\mathcal{A},\mu)$ be a measurabale space and consider the indicator function $1_B$ for $B\in\mathcal{A}$.

Now I want to prove that $1_B\in L_{\mu}^p$ for all $p\geq 1$.

But I have really big problems to show that, because I do not come along with the construction of $L_{\mu}^p$ which is:

$$ L_{\mu}^p:=\mathcal{L}_{\mu}^p/\mathcal{N}_{\mu}^p $$

with for $1\leq p<\infty$

$$ \mathcal{L}{\mu}^p:=\left\{f\colon\Omega\to\overline{\mathbb{R}}\text{ measurable}: \int\lvert f\rvert^p\, d\mu<\infty\right\}, $$ $$ \mathcal{L}_{\mu}^{\infty}:=\left\{f\colon\Omega\to\overline{\mathbb{R}}\text{ measurable}: \exists K > 0 \lvert f\rvert\leq K\text{ a.s.}\right\} $$ and $$ \mathcal{N}_{\mu}^p:=\left\{f\in\mathcal{L}_{\mu}^p: f=0\text{ a.s.}\right\}. $$

So there is an equivalence relation given by $$ f\sim g:\Leftrightarrow f-g\in\mathcal{N}_{\mu}^p $$ and so it is $$ L_{\mu}^p=\left\{[f]: f\in\mathcal{L}_{\mu}^p\right\}. $$ So the elements of $L_{\mu}^p$ are the equivalence classes $F=f+\mathcal{N}_{\mu}^p$ ($f\in\mathcal{L}_{\mu}^p$).

But how can I now show that $1_B\in L_{\mu}^p$ for all $p\geq 1$? I do not see that, because $1_B$ is a function and not an equivalence class!

Can you say me how to show it?

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    What you actually what to show is that there is an equivalence class $[f] \in L_\mu^p$ with $1_B \in [f]$. People are just often sloppy with this, and simply say $1_B \in L_\mu^p$... – fgp Jan 11 '14 at 14:12

1 Answers1

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Show that $[1_B] \in L^p_\mu$. That's the intention.

Since it would be awfully cumbersome to always speak of equivalence classes of functions, people sloppily say that a function $f$ belongs to $L^p_\mu$ instead of saying that its equivalence class modulo equality almost everywhere belongs to $L^p_\mu$.

Sometimes, succinctness of expression is (considered) more important than formal correctness.

Daniel Fischer
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  • It is $[1_B]=1_B + \mathcal{N}{\mu}^p$, right? What do I have to show in order to show that $[1_B]\in L{\mu}^p$? Only that $1_B\in\mathcal{L}_{\mu}^p$? –  Jan 11 '14 at 14:21
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    Yup, $[1_B] = 1_B + \mathcal{N}^p_\mu$. And $1_B\in \mathcal{L}^p_\mu$, i.e. $1_B$ is measurable and has finite integral, is precisely what you need to show. – Daniel Fischer Jan 11 '14 at 14:27
  • $1_B$ is measurable, because $B\in\mathcal{A}$. Now consider $1\leq p<\infty$, then $\int_{\Omega}\lvert 1_B\rvert^p, d\mu=\int_B, d\mu=\mu(B)$. Now this is $<\infty$ only if the measure is finite, right? If it is not finite, it does not have to be smaller than $\infty$? And for $p=\infty$ I have to find a $K>0$ so that $\lvert 1_B\rvert\leq K$ a.s.. I think this is $K=1$, because $1_B\leq 1$ on whole $\Omega$ so the only set where $\lvert 1_B\rvert > 1$ is the emptyset which is a null set. Is that right? –  Jan 11 '14 at 14:34
  • Right, $1_B \in \mathcal{L}^\infty_\mu$ for all $B\in\mathcal{A}$; you have $\lvert 1_B(x)\rvert \leqslant 1$ everywhere (not only a.e.). And $1_B\in \mathcal{L}^p_\mu$ for a $p \in [1,\infty)$ if and only if $\mu(B) < \infty$. – Daniel Fischer Jan 11 '14 at 14:49
  • Anytime when the only set where a property is not valid is the empty set one can leave the a.s. out? I thought, because the emptyset is a null set, one has to add the a.s. –  Jan 11 '14 at 15:00
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    You can omit the a.s. or a.e. if the set ${ x : \lnot P(x)}$ is $\varnothing$. Then $P$ holds everywhere or surely, which is stronger than "almost everywhere/surely". But you can also write $\lvert 1_B(x)\rvert \leqslant 1$ a.s., that is no less correct. – Daniel Fischer Jan 11 '14 at 15:04