Measure Theory: outer regularity With a couple of friends I am working through some of Tao's Introduction to Measure Theory, which is very good so far.  We are stuck on Lemma 1.2.12, Outer Regularity.  It says this.
Let $E\subset \Re^d$ be an arbitrary set.  Then one has
$$m^*(E)=\inf_{E\subset U,U open} m^*(U).$$
($m^*$ is the Lebesgue outer measure)
We have used the function $f(x)=1$ if $x$ is rational and $f(x)=0$ if $x$ is irrational for a lot of examples, and also this related set.
$$E=\{(x,y)| \text{$y=0$ if $x$ is irrational and $0\leq y\leq 1$ if $x$ is rational}\}$$
We think the outer measure of $E$ is $0$ because it is the union of countably many line intervals, each of which has measure $0$ in the plane, and then countable additivity gives a total of $0$.  But we also think the lemma says that $E$ has measure $1$ because any open set $U\supset E$ has to fill in the square, so the inf of all those superset $U$s will be $1$.  Which way is right (I hope one of them is)?
Edit. I'm sorry but I described the set $E$ wrong initially.  I've never asked here before and I'm a little nervous.
 A: In the plane, the horizontal line segment $I_y = [0,1] \times \{y\}$ has Lebesgue outer measure zero. This can be proved straight from the definition. Your set $E$ satisfies $$E \subset I_0 \cup I_1$$ so that $m^*(E) = 0$ too. The open set covering the graph doesn't have to cover the entire square, but only the parts that belong to the graph.

To answer the edited question, define $J_x = \{x\} \times [0,1]$. For every $x$ you can use the outer regularity to show that $m^*(J_x) = 0$. Now countable subadditivity comes into play: since
$$ E = I_0 \cup \bigcup_{x \in [0,1] \cap \mathbb Q} J_x$$ you obtain $$m^*(E) \le m^*(I_0) + \sum_{x \in [0,1] \cap \mathbb Q} m^*(J_x) = 0.$$
A: An open set $U$ that contains $E$ does not have to fill the unit square. Think of the analogous problem for the rationals in $[0,1]$. You can cover the rationals in $[0,1]$ with a union of open intervals that get geometrically smaller: cover the $k$th rational $r_k$ with the interval $$\left(r_k-
\frac\varepsilon{2^k},r_k+\frac\varepsilon{2^k}\right)\quad .$$ The union of these intervals is an open set whose outer measure is at most $2\varepsilon$.
Note: Remember that line intervals of the form $\{r\}\times(a,b)$ are not open sets in the plane; open sets need to have "width" as well as "height". But you can adapt the above argument using long, skinny rectangles.
