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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?

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    Indeed, I found that interesting reading; thank you for the link! It now seems to me that the axioms I've given are equivalent to the Daniell integral, which answers all my questions. (Unless there is something subtle but significant I have missed?).2012-01-13

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