If I define the area of a bounded set $C \subset \mathbb{R}^2$ as the Riemann integral of the indicator function of $C$ over $C$. So $A(C) = \int_{C}\aleph_C$.
Problem: prove or find a counterexample: If $\forall \epsilon > 0$ you can cover the set $C$ with a set $B$ where $A(B)=\epsilon$ then $A(C)=0$.
My attempt:
Assume $A(C) = P > 0$. Then take $\epsilon = P/2$. Let's show that this is a condradiction:
If $A \subset B$ and $\aleph_A,\aleph_B$ indicator functions for $A$ and $B$.
$$A(C) = \int_{A}\aleph_A \leq \int_{A}\aleph_B$$ (since $\aleph_B \geq \aleph_A$ given that $\forall x\in A \Rightarrow x\in B$), then we have $$\int_B\aleph_B = \int_A \aleph_B + \int_{B\setminus A}\aleph_B $$
but $\int_{A\setminus B}\aleph_B \geq 0$ since $\aleph_B(x) \geq 0 , \forall x$
Then we have $$\int_{A}\aleph_B \leq \int_B\aleph_B = A(B)$$ and the statement is proven.
Is my argumentation correct?
The proof is correct except for one detail: it only works under an additional assumption, that $A(C)$ exists. It's not immediately evident that $\chi_C$ is integrable over any rectangle $R \supseteq C$.
The fact, however, remains true without the above assumption. To prove it, fix $\varepsilon > 0$ and take an arbitrary rectangle $R \supseteq C$. Now take $B \supseteq C$ with $A(B) < \varepsilon$. Replacing $B$ with $B \cap R$, we can assume that $B \subseteq R$. (It's not difficult to prove that if $A(B)$ exists and is $< \varepsilon$, then $A(B \cap R)$ exists and is $< \varepsilon$ ).
There is a partition $\mathcal{P}$ of $R$ such that $U(\chi_B, \mathcal{P}) < \varepsilon$. Since $ 0 \leqslant \chi_C \leqslant \chi_B$, we have $0 \leqslant U(\chi_C, \mathcal{P}) \leqslant U(\chi_B, \mathcal{P}) < \varepsilon$.
Hence
$$A(C) = \int \limits_R \chi_C = \inf_{\mathcal{P}} \, U(\chi_C, \mathcal{P}) = 0.$$