False beliefs about Lebesgue measure on $\mathbb{R}$ 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?
 A: 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!
A: 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.
A: 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.
A: 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. 
A: 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 $\leq \sum_1^\infty (1/3)^k = 1/2$.
A: 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.
A: 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.
