My textbook said,
If $\int_E f$ exists then, of course, $-\infty\le\int_E f\le+\infty$. If $\int_E f$ exists and is finite, we say that $f$ is Lebesgue integrable, or simple integrable, on $E$ and write $f\in L(E)$. Thus, $$L(E)=\left\{f:\int_E f \text { is finite}\right\}$$
I think $L(E)$ must be defined by $$L(E)=\left\{f:\int_E f \text { exists and is finite}\right\}$$
Why does the textbook omit existence?
Is $f$ in $L(E)$ or not if $\displaystyle\int_E f$ is finite but oscillating.
Add
Let $f(x)=\sin{x}$. Then $\displaystyle\int_\mathbb{R} f$ is finite, since $\displaystyle\left|\int_\mathbb{R} f(x) dx\right| \le 2 \lt +\infty$. Thus, $f \in L(E)$.
$\displaystyle\int_\mathbb{R} f$ is oscillating ceaselessly. So it should be said not to exist. Thus, $f\notin L(E)$.
I think the fact that $f\notin L(E)$ is surely correct. However, if $L(E)$ is defined by $\displaystyle\left\{f:\int_E f \text { is finite}\right\}$, $f$ is in $L(E)$. !!
If $\int_E f$ is finite then it necessarily exists.
Q: "Is $f$ in $L(E)$ or not if $\displaystyle \int_E f$ is finite but oscillating."
A: If $\int_E f$ is oscillating, then $\int_E f$ doesn't exist. And since it doesn't exist, then it can't be finite (contrapositive of "if finite, then exists"). Consider $\int_\Bbb{R} \sin x \, dx$ as a concrete example. It oscillates between finite values but the integral itself is not finite (because its value doesn't exist). So to answer your question, there is no such thing as a "finite but oscillating" integral. If we rephrase your question to remove "finite but" then the answer is "no, $f$ is not in $L(E)$ in that case."