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Studying I observed that a line is measurable and it's measure is $0$ (you can check it here: Show that a straight line has a lebesgue measure of zero).

Let $\mu(x)$ be a measure. I know that a measurable function can be summable. A function is summable if: $$\int\limits_Xf(x)d\mu(x)=\int\limits_Xf_+(x)d\mu(x)-\int\limits_Xf_-(x)d\mu(x)$$ and both right side integrals are finite, where $f_+(x)$ is the positive part and $f_-(x)$ is the negative one.

For example taking $f(x)=x$, $f: X\rightarrow \mathbb{\bar R}=\mathbb{R}\cup\{\pm\infty\}$, X a set, $\mu$ a measure on $X$: $$\int\limits_\mathrm{R}xd\mu(x)=\int\limits_0^{+\infty}xd\mu(x)-\int\limits_{-\infty}^0xd\mu(x)=x^2/2|_0^{+\infty}-x^2/2|_{-\infty}^0=+\infty-\infty$$ So I conclude that the function $f(x)=x$ is not summable. Since the definition of $L^p$ is the following: $$L^p(\mu)=\{f:X\rightarrow\mathbb{\bar R}:f\ \text{measurable}, ||f(x)||_{L^p(\mu)}<+\infty\}$$ Does it mean that the line $f(x)=x$ is not in $L^p$?

james watt
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  • That's correct. Of course, if you replace $\mathbb{R}$ with any closed, bounded interval $[a,b]$, then $f(x) = x$ is in every $L^p$. – Matthew Leingang Oct 22 '18 at 14:39
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    @MatthewLeingang Checking if $f$ is summable or not does not correspond to checking if $f$ is in $L^1$? What for the other values of $p$? – sound wave Oct 22 '18 at 14:49
  • I think the summable property in some way implies the fact that the norm is finite. Obviously if we restrict our domain on subset of $\mathrm{R}$ the norm even more so will be finite. – james watt Oct 22 '18 at 17:21

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