Let $V$ denote the vector space of functions from $R^n$ to $R$. Define a "bleasure" to mean a linear function $b$ from a subspace of $V$ to $R$ such that:
A) If $0 \leq f$ everywhere, and $b(f)$ is defined, then $0 \leq b(f)$
B) If $0 \leq f \leq g$ everywhere, and $b(g)$ is defined and equal to $0$, then $b(f)$ is defined and equal to $0$ as well
C) If $\sum_{i = 0}^{\infty} f_i(x) = f(x)$ for each $x$, and $0 \leq$ each $f_i$ everywhere, and each $b(f_i)$ is defined, then $b(f)$ is defined and equal to $\sum_{i = 0}^{\infty} b(f)$ if this sum converges, and undefined otherwise.
D) If $g$ is a translation of $f$ (in the sense that there is a constant $c$ such that $g(x) = f(x + c)$), and $b(f)$ is defined, then $b(g)$ is defined and equal to $b(f)$.
E) If $f$ is the indicator function for $[0, 1)^n$, then $b(f)$ is defined and equal to $1$.
For example, one bleasure is the Lebesgue integral.
I have two questions about bleasures:
1) Is there any bleasure which is undefined at some Lebesgue-integrable functions?
2) Is there any bleasure which takes on a different value at a Lebesgue-integrable function than the Lebesgue integral does?