Let $\mu$ be the Lebesgue measure. For $0<p<\infty$, is $\left[\mu(\mathbb{R})\right]^{1/p} = \infty$?
I am trying to construct a function with the property that $\mu\{x: |f(x)|>a>0\}<\infty$ and $f$ is not in any $L^p$, $0<p<\infty$.
So I tried to use the constant function $f:\mathbb{R}\to \mathbb{R}$ with $f(x)=1$ which gives $$\left(\int_\mathbb{R}|1|^pd\mu\right)^{1/p} =\left[\mu(\mathbb{R}\right]^{1/p}.$$ Since $p$ can't be $\infty$, I think $\mu(\mathbb{R})$ is "large" enough?
The title looks like it has some typos.
For the function that you are trying to produce you can define a function with support on $[0,1]$. Therefore all the sets $\{x:\ |f(x)|>a>0\}$ are goind to be of measure $\leq 1$, since they are contained in $[0,1]$.
Lets partition the support into a sequence of disjoint intervals, and the define it in each of them as a constant value.
On the interval $(1/2,1]$ define it as $1$.
On the interval $(1/(n+1),1/n]$ define it as something that growth fast enough with respect to the size of the interval, like $n^{n+1}(n+1)$.
Therefore $\|f\|_p^p=\int_{[0,1]}f^p=\sum_nn^{p(n+1)}(n+1)^{p}\mu((1/(n+1),1/n])=\sum_nn^{p(n+1)}(n+1)^{p}\frac{1}{n(n+1)}$.
For all $0<p<\infty$ this diverges because $n^{p(n+1)}(n+1)^{p}\frac{1}{n(n+1)}\to\infty$.