Infinite Lebesgue integral over all infinite measure subsets? This question is in particular for $\mathbb{R}^+$. What properties must a finite function $f$ have such that $\int_A f d\mu = \infty$ for all A with $\mu(A) = \infty$?
As pointed out previously (when the question was not framed correctly), obviously any constant $f$ will have infinite integral.
 A: It is clearly sufficient if there is some $c>0$ such that
$$
M_{c}:=\left\{ x\in X\,\mid\, f\left(x\right)<c\right\} 
$$
has finite measure. Indeed, if $A\subset X$ has infinite measure,
then so has $A\setminus M_{c}$ and we have $f\left(x\right)\geq c$
on this set. Hence,
$$
\int_{A}f\,{\rm d}\mu\geq\int_{A\setminus M_{c}}f\,{\rm d}\mu\geq c\cdot\mu\left(A\setminus M_{c}\right)=\infty.
$$
We will now show that the condition from above is also necessary,
at least if the space $\left(X,\mu\right)$ satisfies the following
property: For each measurable set $A\subset X$ and each $0\leq\alpha\leq\mu\left(A\right)$,
there is a set $B\subset A$ with $\mu\left(B\right)=\alpha$. This
is in particular fulfilled for the Lebesgue measure, as can be seen
by considering the continuous(!) function $x\mapsto\mu\left(A\cap\left(-\infty,x\right)\right)$
and using the intermediate value theorem).
Assume towards a contradiction that for all $c>0$, the set $M_{c}$
from above has infinite measure. For arbitary $n\in\mathbb{N}$, we
thus have $\mu\left(M_{e^{-n}}\right)=\infty$. By assumption, there
is thus a subset $K_{n}\subset M_{e^{-n}}$ with $\mu\left(K_{n}\right)=n$.
Let
$$
A:=\bigcup_{n\in\mathbb{N}}K_{n}
$$
and note $\mu\left(A\right)\geq\mu\left(K_{n}\right)=n\xrightarrow[n\to\infty]{}\infty$,
which implies $\mu\left(A\right)=\infty$. But we have
$$
\int_{A}f\,{\rm d}\mu\leq\sum_{n}\int_{K_{n}}f\,{\rm d}\mu\leq\sum_{n}e^{-n}\mu\left(K_{n}\right)=\sum_{n}\frac{n}{e^{n}}<\infty.
$$
