In one of our exams it was written without explanation that $f_{n}(x):=\frac{x^n}{x^3+x^4}\chi_{[0,1[} \in \mathcal{L}^1$
But I fail to see how this does not warrant an explanation.
My ideas
$\frac{x^n}{x^3+x^4}$ is continuous on $]0,1[$, $\forall n \in \mathbb N$.
I basically want to find a function $g \in \mathcal{L}^1$ such that $f_{n} \leq g, \forall n \in \mathbb N$
on $]0,1[$ this is not difficult as $\frac{x^n}{x^3+x^4}\chi_{]0,1[}\leq\frac{1}{x^3+x^4}$
But what about at $x = 0$, I mean $f_{n}(0)$ is not defined so how can I create a plausible inequality? I do realize that $\{0\}$ has measure zero but how do I successfuly incorporate that into proving that $f_{n} \in \mathcal{L}^1$
Most likely, the goal is to study the convergence in $\mathcal L^1$ of the sequence $\left(f_n\right)_{n\geqslant 1}$ hence the fact that $f_n$ is not integrable for $n\in\{0;1;2\}$ is not a problem.
For $n\geqslant 3$ and $x\in(0,1)$, $$ 0\leqslant \frac{x^3}{x^3+x^4}\leqslant \frac{x^3+x^4}{x^3+x^4}=1 $$ hence $$ 0\leqslant f_n(x)\leqslant x^{n-3}\leqslant 1 $$ and a bounded function on $(0,1)$ is integrable.