Can the union of lines be an area?

Question

Edit: How come when we measure the area using integrals we get a 'nice' answer, but when we sum up the value ("height") of every positive continuous function on an interval $[a,b]$ the sum: $\Sigma U$ when $U=\{f(x)\mid x\in[a,b[\}$ divergs. Why doesn't the second sum describe the are under the graph? Sorry for the long phrasing below. $$$$ It is known that the sum of any uncountable set $U$ of positive numbers diverges $\Sigma U=\infty$. Because of that, if we try to calculate the area under a continuous graph like $f(x)=x$ at $[0,1]$ the "area" diverges. On the other hand, Riemann sums are working because they are defined by taking the limit of a natural number $n\in \mathbb{N}$ to infinity. I wonder what is the reason that the former method doesn't work. An intuitive definition for area is that the points of a set $A$ are "an area" iff: $$\forall a_0\ \forall \epsilon>0\ \exists a\in A \ \ s.t\mid a_0 - a\mid <\epsilon$$ So I tried to use this definition to prove that any set of lines can't form an area, but it seems false for uncountable sets. What is the real reason we can't calculate area using sum of 1-D lines?

Answer 1

1. the area of a line is 0 so it's not positive, so the sum of the area even of uncountable many lines is 0
2. in measure theory, it is false that an uncountable sum of the area of disjointed sets is the area of the union of the sets (area is the measure here). It's just not in the axiom of the measure.
3. what you are defining as an "area" is a dense set, you can look up the definition of a dense subset of a metric space.
4. a dense set doesn't necessarly have an area, for exemple $\mathbb{Q}^2\subset \mathbb{R}^2$ has measure (or area) 0 and is dense.

Answer 2

You can consider, for each $\theta\in [0, 2\pi]$, the line given by $$\ell_\theta (t)= t(\cos(\theta), \sin(\theta))$$.

The union of that lines is $\mathbb{R}^2$