不可数和的发散可数和-DivergentCountableSumfromUncountableSum

4  

Since $f \ge 0$ then either $f>0$ for only countably many terms, or $f>0$ for uncountably many terms. Then uncountably many must be bigger than some $1/n$ for some $n$. Hence we have the result either way.

因为$ f \ ge 0 $然后要么$ f> 0 $只有相当多的条款，要么$ f> 0 $用于不计其数的许多条款。然后不可否认的是，对于一些$ n $，许多必须大于$ 1 / n $。因此我们得到了结果。

2  

Same result as David's but expressed in a more complicated way.

与大卫的结果相同，但以更复杂的方式表达。

If $f\cdot 1_X$ is $|\cdot|$ measurable there is a sequence of simple functions $s_n$ such that $s_n \le f \cdot 1_X$ and $\int s_n \to \infty$.

如果$ f \ cdot 1_X $是$ | \ cdot | $ measurable，则有一系列简单函数$ s_n $，这样$ s_n \ le f \ cdot 1_X $和$ \ int s_n \ to \ infty $。

If the support of all the $s_n$ is countable, then the union of the supports is countable and the integral over this set is $\infty$.

如果所有$ s_n $的支持都是可数的，那么支持的并集是可数的，并且该集合的积分是$ \ infty $。

If the support of any of the $s_n$ is uncountable, then we have $f \ge s_n \ge \alpha \cdot1_A$ for some uncountable $A$ with $\alpha >0$. Now choose any countable subset of $A$ to get the desired result.

如果任何$ s_n $的支持是不可数的，那么我们有$ f \ ge s_n \ ge \ alpha \ cdot1_A $对于一些不可数的$ A $和$ \ alpha> 0 $。现在选择$ A $的任何可数子集来获得所需的结果。