Final answer:
To prove the integral ∫fdμ with a counting measure is definite, we need to show it yields a specific value. This involves summing the function's values over discrete points in the measure space, which is either finite or infinite.
Step-by-step explanation:
The question pertains to the concept of integration with respect to a counting measure μ on a measure space (ω,F). To prove that the integral ∫fdμ is definite, one would have to show that for any measurable function f, the integral yields a specific finite value or is properly undefined (such as in the case of an integral over a function that is not absolutely integrable).
In the context of a counting measure, which counts the number of elements in a set, the integral of a function essentially sums the values of the function over the discrete points within the measure space, providing a finite or infinite sum depending on the function and the space in question.
To compute the integral of a function f with respect to the counting measure, we sum up the values of f over the elements in the measure space. This gives us a definite value for the integral, since we are summing a finite or countable number of terms.
For example, if f is a function that assigns a value of 1 to each element in the measure space, then ∫fdμ would be the sum of all the elements in the measure space, which is a definite value.