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I have two real functions $f$ and $g$ defined by $ f(x) =\begin{cases} \tfrac1x & \text{if } x \in (0,1] \\ 0 & \text{if } x \in \mathbb R\setminus (0,1] \end{cases} \qquad \text{and} \qquad g(x) = \begin{cases} \tfrac1x & \text{if } x \in (1,\infty] \\ 0 & \text{if } x \in (-\infty,1] \end{cases}.$

I have to calculate $\{p \in [0,\infty] \colon f \in \mathcal L^p(\lambda)\}$ and $\{p \in [0,\infty] \colon g \in \mathcal L^P(\lambda)\}$.

So far, I've thought this: Since $\mathcal L^\infty(\lambda)\subseteq\mathcal L^r(\lambda)\subseteq\mathcal L^0(\lambda)$, where $r \in(0,\infty)$, I only have to calculate $\mathcal L^0(\lambda)$, which is defined like this: $\mathcal L^0(\lambda)={f\in\mathcal M(\mathcal E): \lim_{t\to \infty}\lambda({|f|\ge t})=0}$. I am, however, not at all sure how to calculate this. I would appreciate help a lot.

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    Yes, I'm sorry, I must have mistyped it. I guess it makes a lot more sense now.2012-11-15

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For both functions we can integrate explicitly to determine for which $p\in (0,\infty]$ we have $\int\vert \cdot\vert^p dx<\infty$. Take $f(x)$ for example:

$ \int_{-\infty}^\infty \vert f\vert^p dx=\int_0^1 x^{-p}dx= (-p+1)^{-1}x^{-p+1}\Big\vert_0^1$

so we see that in order for this integral to be finite we need $-p+1>0$ or $p<1$. Similarly for $g$:

$\int_{-\infty}^\infty \vert g\vert ^p dx =\int_1^\infty x^{-p} dx =(-p+1)^{-1}x^{-p+1}\Big\vert_1^\infty$ so now we need $-p+1<0$ or $p>1$.

It should be clear that $f\notin L^\infty$, but that $g$ is.

I'm a little unclear of your definition of $L^0$; typically $L^0$ denotes all measurable functions, but it looks like you're looking at some type of weak $L^1$ norm? If you clarify this, I might be able to suggest an approach.

Edit: For the $\mathcal{L}^0$ case, $g(x)$ should be clear since $\mathcal{L}^\infty\subset\mathcal{L}^0$. For $f(x)$, consider $\{\vert f\vert>t\}$; this is precisely $(0,1/t)$. As $t\rightarrow\infty$, the measure of these sets goes to 0. Thus $f\in \mathcal{L}^0$.

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    Ah, okay. I thought that might would be the case, I just wasn't sure. Thank you.2012-11-15