Show that a straight line has a lebesgue measure of zero

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?

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 $\lambda(\{0\})=0$.

Excellent work. The only thing left to take care of now are vertical lines.