I'm aware that this question has been asked several times and over: however, I remain unsatisfied.
The textbook I'm working with is Real Analysis: Modern Techniques and Their Applications (2nd Edition) by Gerald Folland, with the question I have pertaining to Proposition 2.21.
2.21 Proposition. The set of integrable real-valued functions on $X$ is a real vector space, and the integral is a linear functional on it.
The proof for showing that the set in question is a vector space involves invoking the triangle inequality $|af+bg|\leq|a||f|+|b||g|$: this part makes sense. The question I have is whether the function $\sum_{n=1}^{\infty}f$ will also be integrable? Folland states that a real-valued function $f$ is integrable iff $\int f<\infty$…however, if we define $f(x)=1$, then $\int|\sum_{n=1}^{\infty}f|$ is no longer finite thus $\sum_{n=1}^{\infty}f$ is not integrable. From this we should be able to deduce that the set of all integrable real-valued functions is not closed under addition, hence cannot be a vector space, right? Or am I missing something?
Best Answer
A group or vector space being closed under addition means addition of two elements, or by induction any finite number of elements, but not addition of infinitely many elements. For example, in the additive group of integers, $1+1+1+1+\cdots$ is not in the group, although $1+1+1+\cdots+1$ is.