Exercise: Let $\lambda$ be the Lebesgue measure and let $f:\mathbb{R}\to\mathbb{R}$ be an integrable function. Suppose that we have the function $f_n(x):\mathbb{R}\to\mathbb{R}$ given by $f_n(x) = f(x)(1 + (\sin x)^n)$. Calculate the following limit: $$\lim\limits_{n\to\infty}\int_\mathbb{R}f_nd\lambda$$

What I've tried: If I can show that $f_n$ is measurable and integrable then I can use the dominated convergence theorem. I know that $f_n$ is continuous, which means that if $\int_{-\infty}^\infty\left|f_n(x)\right|dx$ exists as an improper Riemann integral then $f_n$ is integrable. I don't really know how to apply this theorem here though.

Let's assume that $f_n$ is integrable. We have that $\left|f_n\right|\leq 2\left|f\right|$, where $2\left|f\right|$ is integrable. The dominated convergence theorem tells us that we have $$\lim\limits_{n\to\infty}\int_\mathbb{R}f_nd\lambda = \int_\mathbb{R}fd\lambda$$ where $\lim\limits_{n\to\infty}f_n =f$ pointwise. Now, $\lim\limits_{n\to\infty}f(x)(1+(\sin x)^n) = f(x)$, for $x \neq k\pi$, with $k\in\mathbb{N}$. There are however, infinitely many $x\in\mathbb{R}$ for which $\left|\sin x\right| = 1$. So I don't think I can conclude that $$\lim\limits_{n\to\infty}\int_\mathbb{R}f_nd\lambda = \lim_{n\to\infty}\int_{\mathbb{R}\backslash X}f_nd\lambda$$ where $X = \{x\in \mathbb{R}: \left|\sin x\right| = 1\}$.

Question: How do I solve this exercise?



For those $x$ such that $|\sin x|=1$ are of countably many, so they form a set of measure zero, that is, $f_{n}(x)\rightarrow f(x)$ a.e. is satisfied, so Lebesgue Dominated Convergence Theorem still applies.

One may check for Theory of Measure and Integration, J Yeh, page 180, for the assertion:

$|f_{n}(x)|\leq g(x)$ a.e., $f_{n}(x)\rightarrow f(x)$ a.e., $g\in L^{1}$, then $\lim_{n}\displaystyle\int f_{n}(x)dx=\int f(x)dx$.

| cite | improve this answer | |
  • $\begingroup$ Thanks for your reply! Could you explain why $f_n$ is integrable? $\endgroup$ – titusAdam Mar 15 '18 at 9:20
  • $\begingroup$ Because $|f_{n}|\leq 2|f|$ and $2|f|\in L^{1}$, then the integral of $|f_{n}|$ is controlled by $2|f|$ and is then $<\infty$ by integrals comparison rule. $\endgroup$ – user284331 Mar 15 '18 at 9:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.