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## Main Question or Discussion Point

Hi,

Some time ago one of my professors told us about a remarkable theorem, which stated something along the lines of: if a function i takes two arguments, one being another function f, and the other being some region R on which the function f is defined, and this function i satisfies some particular properties (such as linearity in f, if disjoint regions st R1 U R2 = R then i(f, R1) + i(f, R2) = i(f, R), etc) which are similar to the integral, then the function i _must_ be the integral of f.

I later asked him again, and he said it was a basic result in measure theory (which he did not pursue further since it was not a measure theory class). However I have been unable to find this theorem again, partly because I never was quite sure of the exact properties the theorem requires, and also because I never pursued more measure theory. I would really appreciate it if someone could recognize this theorem and either point me to a source, clarify the conditions, or even just tell me what it's called.

Thanks!

Some time ago one of my professors told us about a remarkable theorem, which stated something along the lines of: if a function i takes two arguments, one being another function f, and the other being some region R on which the function f is defined, and this function i satisfies some particular properties (such as linearity in f, if disjoint regions st R1 U R2 = R then i(f, R1) + i(f, R2) = i(f, R), etc) which are similar to the integral, then the function i _must_ be the integral of f.

I later asked him again, and he said it was a basic result in measure theory (which he did not pursue further since it was not a measure theory class). However I have been unable to find this theorem again, partly because I never was quite sure of the exact properties the theorem requires, and also because I never pursued more measure theory. I would really appreciate it if someone could recognize this theorem and either point me to a source, clarify the conditions, or even just tell me what it's called.

Thanks!