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Give an example of a sequence of measurable functions $\{f_n\}$ defined on a measurable set $E \subseteq \mathbb{R}^n$ such that the following strict inequalities hold: \begin{align} \int_E \liminf f_n < \liminf \int_E f_n < \limsup \int_E f_n < \int \limsup f_n. \end{align}

MY ATTEMPT : Take $f_n = \chi_{[n, n+1]}$, then $\liminf f_n = 0$ and $\int_{\mathbb{R}} f_n= 1$, so we have $\int_{\mathbb{R}} \liminf f_n = 0 < 1 = \liminf \int_{\mathbb{R}} f_n$. But $\liminf \int_{\mathbb{R}} f_n = \limsup \int_{\mathbb{R}} f_n$ still holds.

Can anyone give me some examples please?

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  • $\begingroup$ That essentially means we need to find a sequence of function that does not converge. Also, we need to make sure that the sequence of the integral also does not converge. This is the idea. $\endgroup$
    – Boka Peer
    Feb 16, 2020 at 10:41
  • $\begingroup$ Yes, but what are the specific examples? $\endgroup$
    – Eunice
    Feb 16, 2020 at 10:50
  • $\begingroup$ I will let you know if I can construct an example. $\endgroup$
    – Boka Peer
    Feb 16, 2020 at 11:33
  • $\begingroup$ Okay, thank you. :) $\endgroup$
    – Eunice
    Feb 16, 2020 at 12:52

1 Answer 1

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Let $E=[0,3]$ and consider $$ f_n \begin{cases} \chi_{[0,2]} & \text{ if } n \text{ is even},\\ \chi_{[2,3]} &\text{ otherwise}. \end{cases} $$ Then, $$\liminf_{n\to +\infty} f_n = 0 \qquad\text{and}\qquad \limsup_{n\to +\infty} f_n = 1,$$ which implies $$ \int_0^3 \liminf_{n\to +\infty} f_n = 0 \qquad \text{and}\qquad \int_0^3\limsup_{n\to +\infty} f_n = 3. $$ Moreover, $$ \int_0^3 f_n(x)\,dx = \begin{cases} 2 & \text{ if } n \text{ is even},\\ 1 &\text{ otherwise}. \end{cases}$$ Hence, $$ \liminf_{n\to+\infty} \int_0^3 f_n(x)\,dx = 1 \qquad \text{and}\qquad \limsup_{n\to+\infty}\int_0^3 f_n(x)\,dx = 2. $$

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