I'm trying to develop intuition about Lebesgue measure on $\mathbb{R}$ and I'd like to build a list of false beliefs about it, for example: every set is measurable, every set of measure zero is countable, the border of a set has measure zero, etc. Can you help me sharing your experience or with some reference list?
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10$\begingroup$ False belief: a set of positive measure $A$ contains an interval (but by Steinhaus theorem we know that it's the case for $A+A$); take fat Cantors. $\endgroup$– Davide GiraudoDec 24, 2011 at 12:32
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6$\begingroup$ False belief: if a function is continuous almost everwhere, then it is equal almost everywhere to a continuous functions, and vice versa. $\endgroup$– MarkDec 24, 2011 at 12:40
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15$\begingroup$ @David: An easier counter example to "a set of positive measure $A$ contains an interval" is the irrationals. Fat Cantor sets are a better counter example to "a set of of positive measure $A$ is dense somewhere" $\endgroup$– HenryDec 24, 2011 at 12:58
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6$\begingroup$ Counterexamples in Analysis - B. Gelbaum, J. Olmsted (Dover, 2003) would be a good reference here. e.g., they show: there is a measurable non-Borel set; there is a set of measure 0 that is not a countable union of closed sets; $\endgroup$– David MitraDec 24, 2011 at 13:11
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1$\begingroup$ This is not just about $\mathbb{R}$, but still: all $\mathbb{R}^n$'s (with the Lebesgue measure) are isomorphic as measure spaces; there is no "invariance of dimension" (as one might falsely believe) $\endgroup$– user8268Dec 24, 2011 at 13:28
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7 Answers
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False belief: the continuous image of a measurable set is measurable.
A counterexample is provided by the Devil's staircase. Since the image of the Cantor set has full measure, it will have subsets, still measurable, which have non-measurable image. The same function also serves as a counterexample to the following:
False belief: if a continuous function has derivative zero almost everywhere, then it is constant.
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$\begingroup$ Nice! How is it false that "if a continuous function has derivative zero almost everywhere, then it is constant" $\endgroup$– TimDec 25, 2011 at 3:03
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6$\begingroup$ @Tim, the Devil's staircase is continuous on the unit interval and has zero derivative almost everywhere (it is locally constant on the complement of the Cantor set), but it is not constant. $\endgroup$– JDHDec 25, 2011 at 15:42
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False belief: a subset of an interval that is both open and dense has the measure of the interval.
A counterexample is obtained by enumerating the rationals on $[0,1]$ and putting an open interval of length $(1/3)^k$ around the $k$th one. The union of these intervals is clearly dense because it contains a dense set (the rationals) as a subset, and it is clearly open because it is a union of open intervals. But meanwhile, its Lebesgue measure is $\leq \sum_1^\infty (1/3)^k = 1/2$.
answered Dec 24, 2011 at 14:00
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$\begingroup$ In fact, we can choose this open subset of measure $\varepsilon$ where $0<\varepsilon< m(I)$, using the continuity of the map $\varepsilon \mapsto m\left(\bigcup_{j\in\mathbb N}\left(r_j-\varepsilon 2^{-j},r_j+\varepsilon 2^{-j}\right)\right)$. $\endgroup$ Dec 24, 2011 at 14:57
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More Cantor madness:
True belief:
There is a measurable set $A$ in $[0,1]$ such that for any interval $U$ in $[0,1]$, both $A\cap U$ and $A^c\cap U $ have positive measure.
False belief:
The continuous image of a set of measure 0 has measure 0.
answered Dec 24, 2011 at 12:42
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$\begingroup$ Could you explain how you can get such a set $A$ in "true belief" statment? $\endgroup$ Jan 5, 2014 at 11:01
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2$\begingroup$ @seriouslydivergent: Enumerate the intervals with rational endpoints. Inside the first, find two disjoint nowhere dense compact sets $A_1$ and $B_1$ with positive measure. Inside the second interval, find $A_2$ and $B_2$ disjoint from each other and from the previous $A_i$ and $B_i$, and so on. Let $A$ be the union of the $A_i$ and $B$ of the $B_i$: both have positive measure on each interval, and they are disjoint. $\endgroup$– Gro-TsenJan 16, 2016 at 6:14
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False Belief: A nowhere dense subset of $\mathbb{R}$ has measure $0$. (Let me recall that a subset $A$ of $\mathbb{R}$ is said to be nowhere dense if the interior of its closure is empty.)
I leave the explanation as to why this is indeed a false belief as an exercise!
answered Dec 24, 2011 at 15:01
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$\begingroup$ Furthermore, examples of full measure meagre sets: mathoverflow.net/questions/43478/… $\endgroup$ Dec 5, 2013 at 16:48
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Consider the following (true) statement:
If $(I_n)$ is a sequence of subintervals of the unit interval and the sum of their lengths is strictly less than $1$, then the $I_n$ do not cover the unit interval.
False belief: This can be proven just by translating $I_1$ to begin at $0$, translating $I_2$ to start end the end of $I_1$ etc. If this worked, then the same would be true for the unit interval in $\mathbb Q$ where the statement is false.
I obvious can't claim this to be original; I got it from MO.
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True and crazy: There exists a subset $E$ of $[0,1]^2$ which meets every line (horizontal, vertical, or slanted) along a measure zero subset of that line, in fact, along at most two points (i.e., no three points of $E$ are ever aligned), yet such that $E$ does not have measure zero, in fact, $E$ meets every closed set of positive measure in $[0,1]^2$. Also, we can arrange so that $E$ is the graph of a function.
This is due to Sierpiński in 1920. His proof (in French) can be found here ("Sur un problème concernant les ensembles mesurables superficiellement", Fund. Math. 1, 112–115). It's a more or less straightforward transfinite induction with length $2^{\aleph_0}$.
So in particular, you shouldn't believe that $\int_{[0,1]^2} f = 0$ follows from $\int_{[0,1]} f(x,y)\,dx = 0$ for every $y$ and $\int_{[0,1]} f(x,y)\,dy = 0$ for every $x$ (take $f$ to be the characteristic function of this $E$).
Along similar lines, using the Continuum Hypothesis, one can find $E \subseteq [0,1]^2$ which has measure $0$ on each horizontal line and $1$ on each vertical line (in fact, take the graph of any well-ordering of $[0,1]$ with order type $\omega_1$).
Edit: It is worth noting that the pathologic behaviour is due to the resulting sets not being measurable. For measurable sets such things cannot happen due to Fubini's Theorem.
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$\begingroup$ +1. Lovely! Is it consistent with ZF that this can't happen? (I'm sure it is but I don't see the proof just yet.) $\endgroup$ Jan 16, 2016 at 6:32
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$\begingroup$ @NoahSchweber: It is consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, so I really don't know what Lebesgue measure means. But you probably implied ZF+DC, in which case I don't know. In related news, there's a 1980 theorem by Friedman that it's consistent with ZFC the last counterexample I mentioned can't occur. $\endgroup$– Gro-TsenJan 16, 2016 at 6:36
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$\begingroup$ Why doesn't Fubini-Tonelly imply $\int_{[0,1]^2} f = 0$? $\endgroup$ Jan 30 at 9:34
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$\begingroup$ @LordOfNumbers In Fubini-Tonelli, $f$ must be measurable. Look at the edit at the end of this answer. $\endgroup$ Jan 30 at 10:06
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False belief: A simple arc in the plane (i.e., a subset of the plane which is homeomorphic to the interval $[0,1]$) has planar Lebesgue measure zero.
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$\begingroup$ .Interesting. Can you give a reference? $\endgroup$ Jan 16, 2016 at 7:12
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$\begingroup$ @user254665 I'm repeating this old answer. Quoting my comment there: "it follows from a more general theorem of R. L. Moore and J. R. Kline (On the most general plane closed point-set through which it is possible to pass a simple continuous arc, Ann. of Math. (2) 20 (1919), 218-223) that every (homeomorph of the standard) Cantor set in the plane is contained in an arc; apply this to a fat Cantor set in the plane." $\endgroup$– bofJan 16, 2016 at 7:18