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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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In 1), are you asking whether there exists a bleasure that does not contain all Lebesgue-integrable functions in its domain? (I ask because the second "any" could be ambiguous.) –  Jonas Meyer Jan 12 '12 at 4:01
    
Sorry; I am asking whether there exists a bleasure which is missing at least one Lebesgue-integrable function in its domain. Naturally, a bleasure cannot be missing all Lebesgue integrable functions from its domain (e.g., its domain is a subspace of $V$, and thus contains the constantly $0$ function; slightly less trivially, there is condition E)). –  Sridhar Ramesh Jan 12 '12 at 4:04
    
That is what I thought, and I guess it was the only reasonable interpretation, but "any" is sometimes ambiguous. I hope my nitpicky edit doesn't detract from the post. It is an interesting question. –  Jonas Meyer Jan 12 '12 at 4:09
3  
You might be interested in reading about the Daniell integral. –  kahen Jan 12 '12 at 5:51
    
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?). –  Sridhar Ramesh Jan 13 '12 at 10:40

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