The following is a question from ISI MMath PMB.I want to know whether the proof is correct.
Let $f:[0,1]\to [0,\infty)$ be a function.Assume that there is $M\geq 0$ such that $\sum\limits_{i=1}^k f(x_i)\leq M,\forall k\geq 1$ and $\forall x_1,...,x_k\in [0,1]$.Show that $\{x|f(x)\neq 0\}$ is countable.
Proof:Let $A=\{x|f(x)\neq 0\}=\bigcup\limits_{n=1}^\infty\{x:f(x)\geq \frac{1}{n}\}$
Let $A_k=\{x|f(x)\geq \frac{1}{k}\}$ be infinite,for some $k\in \mathbb N$.
Then $A_n$ is infinite for all $n\geq k$.
Take a sequence $(x_n)_{n\geq k}$ such that $f(x_n)\geq \frac{1}{n}$ and $x_n\in A_n$
Then $\sum\limits_{n=k}^\infty f(x_n)$ diverges.
Thus we have a contradiction that assures that each $A_n$ is finite,hence $A$ is countable.
