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Trying to solve Exercise $2.6$ from Real Analysis text by Shakarchi.

Integrability of $f$ on $\mathbb{R}$ does not necessarily imply the convergence of $f(x)$ to $0$ as $x \to\infty$. There exists a positive continuous function f on R so that $f$ is integrable on $\mathbb{R}$, but yet $\limsup_{x\to \infty} f(x) = \infty$.

My question is really not about solving this problem, but what does the $\limsup_{x\to\infty}f(x)$ mean in general? I couldn't find a definition for this case anywhere. This question has an answer for this question, but it still wasn't clear how are open sets for extended reals are defined (possibly those of form $(a,\infty]$ etc.)

CiaPan
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Rias Gremory
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2 Answers2

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If you mimic the definition for sequences, which is $$ \limsup_{n\to\infty}a_n=\lim_{m\to\infty}\sup\{a_n:\ n\geq m\}, $$ you can define $$ \limsup_{x\to\infty}f(x)=\lim_{y\to\infty}\sup\{f(x):\ x\geq y\}. $$ In both cases you can put $\inf$ instead of $\lim$, as the sups decrease.

Martin Argerami
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The characterization I like the best about limit superior of a real-valued function $f$ as $x\to\infty$ is that it is the supremum of the set of all limit points that can be attained by some subsequence $\left(x_n\right)_{n=1}^\infty$. It always exists and you can attain it along some subsequence. I suggest reading the fantastic Wikipedia article on the subject.

Wasradin
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