For a natural number, define and . Consider a partition of unity subordinate to the cover . By summing the with the same in condition (4) of Theorem 3-11, one can assume that there is only one function for each , let it be . Now exists if and only converges. But . So the sum converges if and only if exists.
Take a partition of unity subordinate to the cover where for . As in part (a), we can assume there is only one as in condition (4) of Theorem 3-11. Consider the convergence of . One has where . It follows that the sum in the middle does not converge as and so does not exist.
The assertion that . If not necessrily true. From the hypothesis, we only know the values of the integral of on the sets , but don't know how behaves on other intervals -- so it could be that may not even exist for all To correct the situation, let us assume that is of constant sign and bounded on each set . Then is bounded on each interval and so by Theorem 3-12, the integral in the extended sense is same as the that in the old sense. Clearly, the integral in the old sense is monotone on each interval of , and the limit is just .
The sums and have terms of the same sign and are each divergent. So, by re-ordering the terms of , one can make the sum approach any number we want; further this can be done so that there are sequences of partial sums which converge monotonically to the limit value. By forming open covers each set of which consists of intervals for the sum of terms added to each of these partial sums, one gets covers of . Because is zero outside , one can `fatten' up the covering sets so that they are a cover of the real numbers no smaller than 1 without adding any points where is non-zero. Finally, one can take a partition of unity subordinate to this cover. By using arrangements with different limiting values, one gets the result.