I'm trying to prove that the set $B_a = \{ (x,y) : (|x|+1)^a \le y \le e^{x+y} \le y+10 \}$ is a) Lebesgue-measurable and b) its Lebesgue-measure is positive.
a) the set is the intersection of measurable sets like $ \{ (x,y) : g(x,y) \ge 0\}$ or $ \{ (x,y) : g(x,y) \le 0\}$, where $g(x,y)$ is continuous, so is measurable.
b) can't procede on this one. I sure know that $y\ge(|x|+1)^a \ge 1 $ and that $ln(y)-y \le x \le ln(y+10) - y $. But then?
Let us first consider the inequality: $$e^y \leq y + 10$$ With some elementary analysis, you can show that $y \mapsto e^y-y-10$ is strictly monotonous (increasing) over $\mathbb{R}_+$ and using a calculator you can see that the previous inequality is veryfied over $y \in [0, 2.5]$ (you can choose 2 or 2.3 I just took the two first figures from my calculator, the only zero of the function over $\mathbb{R}_+$ being approximately $\approx 2.527963$). Okay...
Now you know that this other inequality is always verified over $\mathbb{R}$: $$y \leq e^y$$ With a little bit of trivial analysis you can show that $\exists \epsilon_1 \in \mathbb{R}_+^*$ such that the following property holds: $$\forall y \in [0,2] \quad e^{\epsilon_1} \times e^y \leq y+10$$ (you can probably find a value for it, but I'm too lazy for that, sorry) meaning the following inequality holds for $y$ over $[0,2]$: $$y \leq e^{\epsilon_1} \times e^y \leq y+10$$
And then, last step before we conclude, with again, a little bit of analysis, you can show that $\exists \epsilon_2 \in \mathbb{R}_+^*$ such that $$(1+\epsilon_2)^{|a|} \leq 1.5 $$ (for what ever value of $a$ you have) Now defining $\epsilon := \mathrm{Min}(\epsilon_1,\epsilon_2)$; you can easily verify that: $$\forall y \in [1.5,2] \quad (1+\epsilon)^{|a|} \leq y \leq e^{\epsilon} \times e^y \leq y+10$$ And then it follows that: $$\forall x \in [0, \epsilon] \quad \forall y \in [1.5,2] \quad (1+|x|)^{|a|} \leq y \leq e^{x+y} \leq y+10$$ Since $\epsilon > 0$ the set $[0,\epsilon] \times [1.5,2]$ is of measure $\frac{\epsilon}{2}$ i.e. has a strictly positive measure. But we have shown that this set is contained in your original set, so your original set has a measure of at least that (i.e. a positive measure too).