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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

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    $\begingroup$ If $x$ is a point on the real line, then for any $\epsilon >0$, we have $x\in (x-\epsilon,x+\epsilon)$ hence $\lambda(\{x\})\leq 2\epsilon$. This argument only uses monotonicity of $\lambda$ and a property of real numbers. Your argument requires the theorem, if $A_n \downarrow A$ then $\lambda(A_n) \downarrow \lambda(A)$ where the first down arrow means $A_n \supset A_{n+1}$ and $A=\cap A_n$ and the second downarrow means $\lambda (A_n) \geq \lambda(A_{n+1})$ and limits to $\lambda(A)$. $\endgroup$ Jan 12, 2018 at 17:03

3 Answers 3

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Another possibility:

closed $\implies$ Borel measurable $\implies$ Lebesgue measurable.

and

$$ \forall\epsilon > 0:\, \{x\}\subset(x−\epsilon,x]\implies \forall\epsilon > 0:\, 0\le\lambda(\{x\})\le\lambda((x−\epsilon,x]) = \epsilon\implies\lambda(\{x\}) = 0. $$

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That could work, but you're going to need to use some property of measures to show that the limit argument at the end goes through. For example, one that could help is that for a decreasing collection of sets $A_i$ of finite measure $\mu$, $\mu(\bigcap_{i=1}^\infty A_i) = \lim_{i \to \infty} \mu(A_i)$ (for a reference, e.g., Bass's Real Analysis for Graduate Students Proposition 3.5). If you haven't already proven properties like this, you might want to think a bit about why this is true.

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One way to make your argument work is to consider a countable sequence of Borel sets (in which at least one set has finite $\lambda$-measure) that decreases to $\{x\}$:

Let $A_n \mathrel{:=} \left(x-1/n, x\right]$ for all $n \in \mathbb N_{\geq1}$ and $x \in \mathbb R$.
Clearly, $A_n \in \mathcal B(\mathbb R) \;\forall n \in \mathbb N_{\geq1}$, and $A_n \downarrow \{x\}$ as $n \to \infty$, i.e. $A_1 \supset A_2 \supset \ldots$ and $\bigcap_{n=1}^\infty A_n = \{x\}$.
Continuity from above gives $\lambda(\{x\}) = \lambda(\bigcap_{n=1}^\infty A_n) = \lim_{n \to \infty} \lambda(A_n) = \lim_{n \to \infty} 1/n = 0.$

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