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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?

Thanks!

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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 $\lim_{n}\displaystyle\int f_{n}(x)dx=\int f(x)dx$.

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  • $\begingroup$ 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

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