Suppose $E \subset \mathbb{R}$ has infinite measure, and $f\in L^1(E)$, Is this true that $f \in L^{\infty}(E)$ necessarily?

Question

Suppose $E \subset \mathbb{R}$ has infinite measure, and $f\in L^1(E)$,

Is this true that $f \in L^{\infty}(E)$ necessarily ?

I could not find a counterexample so far and it is a useful fact if it is true(sometimes to be able to use Holder's inequality)

Thanks for your help

Answer 1

Take a function defined on $[1/(n+1),1/n]$ as $\sqrt{n}$ on the interval $[0,1]$, and zero otherwise.

Answer 2

Take $E=(0,\infty)$ and let $$f=\begin{cases}\frac{1}{\sqrt{x}} & x\in [0,1]\\ 0&\mathrm{o.w.}\end{cases}$$ Then $f\in L^1(E)$, but not in $L^\infty(E)$.

Answer 3

Actually, taking $$f(x):=\sum_{j=1}^\infty2^{-j}\frac 1{\sqrt{|x-r_j|}},$$ where $(r_j)_{j=1}^\infty$ is an enumeration of rationals, we can see that

• $f$ is integrable;
• if $I\subset (0,1)$ is an interval, then $f$ is unbounded on $I$.