OSA: Theory of Integration

Question 1

Decide which of the following claims are correct. Note: By \(\lambda_d\) we denote the Lebesgue-measure on \(\mathbb{R}^d\).

claim	correct?	explanation
If \(\mathcal{A}\) is a \(\sigma\)-algebra on a set \(X\), and \(A,B \subseteq X\) with \(A,B \not\in \mathcal{A}\), then \(A \cup B \not\in \mathcal{A}\).	yesno	Choose, for example, any non-measurable set and its complement for \(A\) and \(B\).
Let \(\mathcal{A}\) be the powerset of \(\mathbb{N}\) (the set of natural numbers) and \[\mu(A) := \begin{cases}1&\text{if }A\text{ is infinite}\\0&\text{if }A\text{ is finite}\end{cases}\] Then \(\mu\) is an outer measure on \(\mathbb{N}\).	yesno	\(\mu\) is monotone and satisfies \(\mu(\emptyset) = 0\), but it is not coutable subadditive, since for example \(1 = \mu(\mathbb{N}) = \mu(\bigcup_n \{n\}) > \sum_n \mu(\{n\} = 0\).
Let \(A \subseteq \mathbb{R}^2\) be compact for some \(d \in \mathbb{N}\), and \(f : \mathbb{R}^d \to \mathbb{R}^d\) continuous. Then \(\lambda_d\big(f(A)\big) < \infty\).	yesno	The image of a compact set under a continuous function is again compact, therefore \(f(A)\) is in particular bounded and measurable and its Lebesgue-measure is finite.
Every measure is an outer measure.	yesno	By definition a measure is coutably additive, which implies countable subadditivity.
Let \(f:\mathbb{R}^2 \to \mathbb{R}^2\) be measurable. Then \(\int_{\mathbb{R}^2} f(x,y)\,\mathrm{d}\lambda_2(x,y) = \int_{\mathbb{R}}\int_{\mathbb{R}} f(x,y)\,\mathrm{d}\lambda_1(x)\mathrm{d}\lambda_1(y)\).	yesno	Note that in Fubini's Theorem we ask \(f\) to be not just measurable but integrable, i.e. \(\int \|f\| \,\mathrm{d}\lambda_2 < \infty\). Indeed, \(f(x,y) = \frac{x^2-y^2}{(x^2+y^2)^2}\) is a counterexample to the claim in question.

Question 2

Compute the following limit (integration is with respect to Lebesgue-measure): \[ \lim_{n \to \infty} \int_{[0,n]} e^{-\sqrt{x^2+n^{-2}}}\,\mathrm{d}x \]

Your answer: