Sum of a zero-measure sets

Question

Maybe a very basic question, but I don't have a specific measure theory background:

If $1\ge\sum_{i=0}^\infty\mu(A)$, where $\mu$ is the standard Lebesgue measure (on a phase space), does it imply that $\mu(A)=0$?

Answer 1

If not, say, $\mu(A):=\alpha>0$, then $\displaystyle\sum_{i=0}^{N}\mu(A)=(N+1)\alpha$ and $(N+1)\alpha\leq 1$, or $(N+1)\leq\dfrac{1}{\alpha}$ for all $N=1,2,...$, then choose a large $N$ such that $\dfrac{1}{\alpha}<N$, then we end up with a contradiction that $N+1<N$.

Answer 2

That has nothing to do with measure theory but simple analysis. For $c\in\mathbb R$ you have $$ \sum_{i=0}^\infty c = \begin{cases} \infty & c>0\\0 & c=0\\-\infty & c<0\end{cases}. $$