TITLE: Show that a straight line has a lebesgue measure of zero
QUESTION [12 upvotes]: Prove that under Lebesgue Measure on $\mathbb{R}^{2}$ every straight line has a measure of zero.
My try:
Let A = $\{ (x,f(x)), x \in \mathbb{R} \}$ be the set of all points lying on the straight line $y = f(x)$.
Show that $\lambda(A) = 0$.
Let $\{ (a_{i}, b_{i})\}$ be a sequence covering $\mathbb{R}$ with $ b_{i} - a_{i} = 1$ $ \forall i \in \mathbb{N}$.
Define $K_{i} = \{ (x,f(x) - \frac{\epsilon}{2^{i}})\times (x,f(x) + \frac{\epsilon}{2^{i}}), x \in (a_{i}, b_{i}) \} $
Then K = $ \cup_i K_{i}$ covers A and 
$\lambda(K) \leq \sum_{i}(b_{i} - a_{i}) \frac{2\epsilon}{2^{i}} = 2\epsilon \rightarrow 0, \epsilon \rightarrow 0$.
Since $K$ covers $A$,
$\lambda(A) \leq \lambda(K)$  
Is this ok or I have I missed something or done something wrong?

REPLY [8 votes]: Great work. Here is an alternative that could make things slightly (just slightly) easier to write down.
First observe that there exists an affine translation of $\mathbb{R}^2$ which preserves measure (a rotation + a translation), and takes your straight line to 
$$
L=\{(x,0)\;;\;x\in\mathbb{R}\}.
$$
Now your line has the same measure as $L$.
Now by Fubini
$$
\lambda(L)=\int_{\mathbb{R}^2}1_Ldxdy=\int_{x\in \mathbb{R}} \left(\int_{y\in\mathbb{R}}1_L(x,y)dy \right)dx=\int_{x\in\mathbb{R}}\left(\int_{y\in\{0\}}1dy \right)dx=0
$$
since $\lambda(\{0\})=0$.