One is given two intervals $I_{a-\epsilon,b+\epsilon}$, $I_{a,b}$ of $\mathbb{R}^n$, and is asked to show that $\lambda(I_{a-\epsilon,b+\epsilon}) - \lambda(I_{a,b}) \leq c\epsilon$ for some constant $c$., where $\lambda(I_{a,b}) = \prod_{i = 1}^n (a_i - b_i)$.
Now, doing some direct calculations I end up with $\lambda(I_{a-\epsilon,b+\epsilon}) - \lambda(I_{a,b}) = \sum_{i=1}^n (2\epsilon)^i C_i$ where the $C_i$'s are constants not dependent on $\epsilon$. The only assumption made on $\epsilon$ is as usual $\epsilon > 0$. If one assumes $\epsilon < 1$ then we get the required inequality. However assuming $\epsilon < 1$ is not being general. Am I missing some (simple) approximation of products?
edit: $I_{a-\epsilon,b+\epsilon} = \{x \in \mathbb{R}^n \mid a_i - \epsilon < x_i < b_i + \epsilon$, $i = 1,\dots,n\}$
Take $n=2$, $a=(0,0)$ and $b=(1,1)$. Then $$I_{a-\varepsilon,b+\varepsilon} = (-\varepsilon,1+\varepsilon) \times (-\varepsilon,1+\varepsilon)$$
implies
$$\lambda( I_{a-\varepsilon,b+\varepsilon}) -\lambda(I_{a,b})= (2\varepsilon+1)^2-1.$$
Since this function grows as $\varepsilon^2$ as $\varepsilon \to \infty$, we cannot expect that there exists a constant $c>0$ such that
$$\forall \varepsilon>0: \lambda( I_{a-\varepsilon,b+\varepsilon}) -\lambda(I_{a,b}) \leq c \varepsilon.$$
Consequently, the estimate
$$\lambda( I_{a-\varepsilon,b+\varepsilon}) -\lambda(I_{a,b}) \leq c \varepsilon$$
can only hold true for $\varepsilon \leq \varepsilon_0$ where $\varepsilon_0$ is some fixed constant.