Relation between Riemann integral and Lebesgue integral on an unbounded interval

Question

The original problem is to compute $\lim_{n \to \infty} \int_{\mathbb{R}}\frac{\cos^n x}{1+x^2} dm(x),$ where $m$ is the Lebesgue measure. I think this can be done by observing that $f_n(x):= \frac{\cos^n x}{1+x^2}$ is bounded above by $g(x) := \frac{1}{1+x^2}$, which is integrable on $\mathbb{R}$, and also that $f_n(x)$ converges to $0$ a.e. Hence, by the dominated convergence theorem, the answer is $0$.

My question is whether the following limit of Riemann integral (instead of Lebesgue integral) also makes sense and is equal to the above : $$\lim_{n \to \infty} \int_{-\infty}^{\infty}\frac{\cos^n x}{1+x^2} dx.$$

Here is my attempt: Using $f_n(x) \leq g(x)$ for all $x$ and $\int_{-\infty}^{\infty}g(x) dx = \pi$, we can say that the improper (Riemann) integral exists (and is less than or equal to $\pi$), which depends on $n$. But it is not clear to me why the limit exists.

Another thing I know is the following result:

Let $f$ be a bounded function on a compact interval $[a, b]$. If $f$ is Riemann integrable on $[a, b]$, then $f$ is Lebesgue integrable on $[a, b]$, and $$\int_a^b f(x) dx = \int_{[a, b]} f(x) dm(x).$$

But this result requires a compact interval, which is not the case here.

To summarize, could anyone give a rigorous explanation of how to show $$\lim_{n \to \infty} \int_{\mathbb{R}}\frac{\cos^n x}{1+x^2} dm(x) = \lim_{n \to \infty} \int_{-\infty}^{\infty}\frac{\cos^n x}{1+x^2} dx,$$ if it is indeed true? Thank you.

Answer 1

We can easily extend the theorem you quoted to handle unbounded intervals:

Theorem $\star$

Suppose $f:\Bbb{R}\to\Bbb{C}$ is a function with the following properties:

• $f$ is improperly Riemann-integrable on $\Bbb{R}$ (i.e for every $-\infty<a<b<\infty$, the restricted map $f|_{[a,b]}:[a,b]\to\Bbb{R}$ is properly Riemann-integrable, and the limit $\lim\limits_{\substack{b\to\infty\\a\to-\infty}}\int^b_af(x)\,dx$ exists in $\Bbb{R}$; this limit being denoted $\int_{-\infty}^{\infty}f(x)\,dx$).
• $f$ is Lebesgue-integrable on $\Bbb{R}$: $\int_{\Bbb{R}}|f|\,dm<\infty$ (Lebesgue-measurability of $f$ is guaranteed by the previous condition)

Then, the two integrals coincide: \begin{align} \int_{\Bbb{R}}f\,dm&=\int_{-\infty}^{\infty}f(x)\,dx. \end{align}

The proof is immediate from the theorem you quoted and dominated convergence: \begin{align} \int_{\Bbb{R}}f\,dm&=\lim_{n\to\infty}\int_{[-n,n]}f\,dm\tag{DCT}\\ &=\lim_{n\to\infty}\int_{-n}^nf(x)\,dx\tag{your quoted theorem}\\ &=\int_{-\infty}^{\infty}f(x)\,dx, \end{align} where the final equality is because $f$ is assumed to be improperly Riemann-integrable.

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Next, you write

Here is my attempt: Using $f_n(x) \leq g(x)$ for all $x$ and $\int_{-\infty}^{\infty}g(x) dx = \pi$, we can say that...

Well, this by itself is not good enough. You need to say $|f_n(x)|\leq g(x)$ for all $x$; the absolute values are important (and I assume you know why this inequality is sufficient to guarantee the improper Riemann-integrability of $f_n$ on $\Bbb{R}$).

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Finally, we know that for each $n\in\Bbb{N}$, the function $f_n$ satisfies the conditions of theorem $(\star)$. So, $\int_{-\infty}^{\infty}f_n(x)\,dx=\int_{\Bbb{R}}f_n\,dm$. By the dominated convergence theorem (applied to the right side), we know the limit exists and equals $0$. Therefore, \begin{align} \lim_{n\to\infty}\int_{-\infty}^{\infty}\frac{\cos^n(x)}{1+x^2}\,dx&=\lim_{n\to\infty}\int_{\Bbb{R}}\frac{\cos^n(x)}{1+x^2}\,dm(x)=0. \end{align} Hopefully it's clear which are (improper)Riemann integrals and which are Lebesgue integrals.