I want to show that the Lebesgue measure of a single point $x \in \mathbb{R}$ is $0$ i.e. $\lambda(\{x\})=0$.
I have thought of something, but I'm not sure if this is correct.
ATTEMPT:
We choose $x \in \mathbb{R}$ and $\epsilon > 0$ arbitrarily. We know per the definition of the Lebesgue measure, that $\lambda( (x-\epsilon,x]) = x- (x-\epsilon)=\epsilon \leq \epsilon$.
Thus, we can conclude, that $\lambda( (x-\epsilon,x])\rightarrow0$ when $\epsilon \rightarrow 0$. We know that $(x-\epsilon,x] \rightarrow \{x\}$ when $\epsilon \rightarrow 0$. Can we conclude now that $\lambda(\{x\})=0$?
Thanks for your time,
K. Kamal
Best Answer
Another possibility:
closed $\implies$ Borel measurable $\implies$ Lebesgue measurable.
and
$$ \forall\epsilon > 0:\, \{x\}\subset(x−\epsilon,x]\implies \forall\epsilon > 0:\, 0\le\lambda(\{x\})\le\lambda((x−\epsilon,x]) = \epsilon\implies\lambda(\{x\}) = 0. $$