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I have a quite dumb question. Is the empty set measurable? say with respect to the standard measure.

I totally acknowledge intuitive explanations. Thanks,

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  • $\begingroup$ To be honest, I find the result quite counter-intuitive. For example, if $S$ and $T$ are distinct measurable non-empty sets of real numbers with $S$ a subset of $T$, then we can say that $\mu(S) < \mu(T)$. This is intuitive - if you have a line and then you cut a bit off then the resulting line is shorter. However if you remove the non-empty restriction then this breaks down - let $S = \emptyset$ and $T = \{0\}$, then $S$ is a distinct subset of $T$ but $\mu(S) = \mu(T)$. However, despite my intuition, the empty set is indeed considered to be measurable, as answers have indicated. $\endgroup$
    – Kidburla
    May 11 '15 at 17:53
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    $\begingroup$ @Kidburla The inequality you wrote down does not hold whenever $S$ is a subset of $T$, and both are non-empty. For example, in the Lebesgue ("length") measure, both the closed interval $[0,1]$ and its subset, the open interval $]0,1[$, have measure $1$. Another example would be the sets $\{0\}\subset\{0,1\}$, both with measure $0$. Intuitively, this is because you removed a set "too small" to change the length. $\endgroup$
    – Andrea
    May 11 '15 at 19:13
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Intuition

Given a set $X$, to avoid paradoxes such as the Banach Tarski paradox , there is a need to define which subsets are measurable (and as a consequence which ones aren't), and then define a function $\mu$, called a measure, from this set $\Sigma$ of subsets of $X$ to the non-negative reals. How to choose?

You want the measure to mirror our intuition of volume, or length, generalising it to higher dimensions and/or to non-metric spaces. In particular, you want the whole set $X$ to have a measure, so that $X \in \Sigma$. Another property you want is that if a subset $A\in \Sigma$ has a measure, you want its complement $X/A$ to have measure $\mu(X)-\mu(A)$. These two properties are necessary (but not sufficient) for $(X,\Sigma,\mu)$ to be a measure space, i.e. as space that generalises our intuition of volume without logical inconsistencies.

Answer

For any measure space $(X,\Sigma,\mu)$, the empty set $X / X \equiv \emptyset$ needs to be a measurable set, and its measure is invariably $\mu(X)-\mu(X) = 0$.

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    $\begingroup$ Good explanation/intuition, but just in case this causes any confusion, it might be worth clarifying that your argument only applies to finite measure spaces (ie, with $\mu(X) < \infty$). It is also possible to have $\mu: \Sigma \to \mathbb{R}^{+}_0 \cup \{ \infty \}$. The argument can easily be rectified by taking some $A \subset X$ with $\mu(A)$ finite, then $\mu(A) + \mu(\emptyset) = \mu(A \cup \emptyset) = \mu(A)$ [the first equality comes as $A \cup \emptyset = \emptyset$]. The result $\mu(\emptyset) = 0$ follows by subtracting $\mu(A)$ from both sides - but this requires $\mu(A)$ finite. $\endgroup$
    – David E
    May 11 '15 at 15:11
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Measures are defined on $\sigma$-algebras (containing the measurable sets) and any $\sigma$-algebra contains the empty set by definition.

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Any $\sigma$-algebra contains $\emptyset$ and for any measure $\mu$ on the $\sigma$-algebra, $\mu(\emptyset) = 0$.

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Assume there is a set $A$: $\mu(A)<\infty$. Let $\mu(A)=k$. Then $\begin{equation}\mu(\emptyset)=\mu(A-A)=\mu(A)-\mu(A)=k-k=0.\end{equation}$

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