An example of a function $f ∈ L_1(\mathbb{R})$ such that $\limsup_{x→\infty} f(x) = \infty$ I have come with an example of such a function which I'm a little unsure of. My course doesn't cover the Lebesgue integration very well, so I am trying to clarify my doubts here.
My function is $f(x)= n$ whenever $x=n\in\mathbb{N}$ and $0$ otherwise. Clearly, $\limsup_{x→\infty} f(x) = \infty$ but how do I show this function is in $L_1(\mathbb{R})$ with rigour? Any ideas?
 A: In $f(x)$, u defined, $f(x)≠0$ in a set of measure $0$. So wont affect the integration and hence $L_1(R)$.
A: Hint
You can even find a continuous map with such properties, i.e. $f(n)=n$, $f(x)=0$ for $$x \in \mathbb R \setminus \bigcup_{n\in \mathbb N} [n-1/n^3,n+1/n^3]$$ and for $x \in [n-1/n^3,n+1/n^3]$, the shape of $f$ is a triangle.
You have
$$\int_{\mathbb R} f = \sum 1/n^2$$ which converges.
A: The support of your function is denumerable so it has Lebesgue measure 0 (by $\sigma-$additivity) therefore $\int f =0$. In fact your function is null a.e..
So your formulation using $\limsup$ is not really correct. When you speak of $f$ as element of $L^1$ you refer to a class of function with respect to the equivalence relation of being equal a.e. (i.e. apart from a set of Lebegue measure 0).
A: Define $u(x)=\Sigma_{k=1}^{\infty}ku_k(x)$, $u_k(x)$ is characteristic function of set $[k-\frac{1}{2k^3},k+\frac{1}{2k^3}]$. So we have $\int_{-\infty}^{+\infty}|f(x)|dx <\int_{-\infty}^{+\infty}|u(x)|dx=\Sigma_{k=1}^{\infty}\frac{1}{k^2}=\frac{\pi ^2}{6}$, which means $f\in L^1(\mathbb R)$
