Example of integrable function which is nowhere $p$-integrable

Question

Can we find an example of a function $g\in {\rm L}^1(0,1)$ which satisfies: for arbitrary $1<p<+\infty$ and for arbitrary measurable subeset $A\subseteq (0,1)$ such that $\lambda(A)>0$ it holds that $g\notin {\rm L}^p(A)$? I am aware of the fact that we can construct ${\rm L}^1(0,1)$-functions which are nowhere bounded, i.e., $g\notin {\rm L}^{\infty}(A)$ for arbitrary measurable subeset $A\subseteq (0,1)$ such that $\lambda(A)>0$ (I can't recall the actual textbook where I read it), but I do not know if we can violate $p$-integrability everywhere and for every $p>1$.

Answer 1

No, the statement you say you read is false. (I suspect that what you actually read was that in some context there exists $f\in L^1$ such that there is no non-empty open set $A$ on which $f$ is bounded...)

If $\mu(X)>0$ and $f\in L^1(\mu)$ then there exists $A$ with $\mu(A)>0$ such that $f$ is bounded on $A$.

(Of course if $\mu(X)=0$ there are no sets of positive measure to begin with, hence the restriction.)

Proof: For $n=1,2\dots$ let $$A_n=\{x:|f(x)|\le n\}.$$We need to show that some $A_n$ has positive measure. If on the other hand $\mu(A_n)=0$ for every $n$ then for every $n$ we have $|f|>n$ almost everywhere, so $$\int|f|\,d\mu>n\mu(X).$$Hence $\int|f|=\infty$, contradiction.

There is an example of the sort you ask about for open sets instead of measurable sets. First concoct $g\in L^1(\Bbb R)$ such that for every $\delta>0$ and every $p>1$ you have $$\int_{-\delta}^\delta|g|^p=\infty.$$Now say $(x_n)$ is a countable dense set and let $$f(x)=\sum\frac1{n^2}g(x-x_n).$$Then $\int_A|f|^p=\infty$ for any non-empty open set $A$.