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I am stuck on two problems in introductory measure theory on the convergence theorems (monotone convergence theorem and dominated convergence theorem).

The exercise asks to compute the limit as $n\to\infty$ of the following integrals.

$$\int_{\mathbf{R}^+}\frac{ne^{-nx}}{\sqrt{1+n^2x^2}}\,dx$$

$$\int_{\mathbf{R}}\frac{e^{-x^2}}{2\cos(\frac{x}{n})-1}\mathbf{1}_{\{3|\cos\left(\frac{x}{n}\right)|\geqslant2 \}}\,dx$$

To apply the dominated convergence theorem, we have to show that we have a sequence $f_n$ of Lebesgue-integrable functions, with $f_n\to f$ $\mu$-almost everywhere, and a Lebesgue-integrable function $g$ with $|f_n|\leq g$ for all $n$ $\mu$-almost everywhere. Then we can interchange limit and integral.

My thoughts:

$$\frac{ne^{-nx}}{\sqrt{1+n^2x^2}}=\frac{e^{-nx}}{\sqrt{\frac{1}{n^2}+x^2}}\xrightarrow{n\to\infty} 0,$$

hence we cannot apply the monotone convergence theorem. My hope goes to the dominated convergence theorem, so I try to look for a function that dominates. The bound $|f_n(x)|\leq \frac{1}{x}$ isn't helpful, since $\frac{1}{x}$ isn't Lebesgue-integrable. I try $|f_n(x)|\leq ne^{-nx}$, but I don't see how to proceed..

For the second one, all functions are bounded by $3e^{-x^2}$ by using the condition of the indicator, which is Lebesgue integrable. But I don't see what the limit if of this sequence of functions.. Given what the graph below looks like, I think it must be $e^{-x^2}$, but I don't see how to prove this.

Any help is appreciated.

enter image description here

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You cannot deal with the first one using the dominated convergence theorem. With $$f_n(x) = \frac{ne^{-nx}}{\sqrt{1 + n^2x^2}}$$ for $x > 0$ we have $f_n(x) = n\cdot f_1(nx)$, and by the change-of-variables formula we have $$\int_{\mathbf{R}^+} f_n(x)\,dx = \int_{\mathbf{R}^+} f_1(x)\,dx > 0$$ for all $n > 0$, while as you found $\lim_{n \to \infty} f_n(x) = 0$ for all $x > 0$. If the dominated convergence theorem were applicable, the limit would have to be $0$ since that is the integral of the pointwise limit.

For the second one it suffices to note that since $\lim_{n \to \infty} \frac{x}{n} = 0$ for every $x \in \mathbf{R}$ and the cosine is continuous with $\cos 0 = 1$, every $x$ lies in $$A_n = \biggl\{ x \in \mathbf{R} : 3\Bigl\lvert \cos \Bigl(\frac{x}{n}\Bigr)\Bigr\rvert \geqslant 2\biggr\}$$ for all sufficiently large $n$. What $n$ are sufficiently large of course depends on $x$, but that doesn't matter. And thus $$\lim_{n \to \infty} \frac{e^{-x^2}}{2\cos \bigl(\frac{x}{n}\bigr) - 1}\mathbf{1}_{A_n}(x) = \lim_{n \to \infty} \frac{e^{-x^2}}{2\cos \bigl(\frac{x}{n}\bigr) - 1} = \frac{e^{-x^2}}{2\lim_{n\to \infty} \cos \bigl(\frac{x}{n}\bigr) - 1} = \frac{e^{-x^2}}{2\cdot 1 - 1} = e^{-x^2}$$ for all $x$.

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