Let $\Omega$ be a bounded domain in $\mathbb{R}^N$. Let $f(x)=|x|^\alpha$. Then $f\in L^1(\Omega)$ if $\alpha>-N$.
The above fact seems to hold for the following reason.
Since $\Omega$ is bounded, there exists $R>0$ such that $\Omega\subset B(0,R)$. Then, we have \begin{align*} I&=\int_{\Omega}f(x)dx\\ &\leq\int_{B(0,R)}|x|^\alpha dx\\ &=\frac{r^{\alpha+N}}{\alpha+N}\Bigg|_{0}^{R}\\ &=\frac{R^{\alpha+N}}{\alpha+N}, \end{align*} where $\alpha+N>0$.
Then we get $f\in L^1(\Omega)$.
Kindly inform me, if the above argument seems fine with you.
Thanks.
The idea behind the proof is correct, but I believe there is a computation error (off by a scalar multiple). Note that the $N$-dimensional ball is a union of concentric $(N-1)$-dimensional spherical shells, hence we have that $$\frac{dV_N(r)}{dr}=S_{N-1}(r)=r^{N-1} S_{N-1}(1) \tag{$\ast$}$$ where $V_n(r)$ and $S_{n}(r)$ are the volume and surface area of the $n$-ball and $n$-sphere of radius $r$ respectively. We use hyperspherical coordinates $(r,\varphi_1,\varphi_2,\dots,\varphi_{N-1})$ to evaluate the integral. Note that one can simply plug in $dV=r^{N-1} S_{N-1}(1)~dr$ (see the earlier computation in $(\ast)$, and note the abuse in notation) due to radial symmetry of the integrand. Hence, for $R>0$ such that $\Omega\subset B(0,R)$, we have that $$I=\int_{\Omega} f(x)~dV\leq \int_{B(0,R)} |x|^{\alpha}~dV=\int_{0}^R r^{\alpha}\cdot r^{N-1} S_{N-1}(1)~dr=S_{N-1}(1)\cdot \frac{R^{\alpha+N}}{\alpha+N},$$ which is finite for $\alpha>-N$, hence $f\in L^1(\Omega)$.
In case you are interested, an explicit formula for $S_n(1)$ for $n\in \mathbb{N}$ is given by $$S_n(1)=\frac{2\pi^{(n+1)/2}}{\Gamma((n+1)/2)},$$ where $\Gamma$ is the gamma function.
Note: If radial symmetry does not hold, the volume element is given by (see the first link) $$dV= r^{N-1}\sin^{N-2}(\varphi_1)\sin^{N-3}(\varphi_2)\cdots \sin(\varphi_{N-2})\, dr\,d\varphi_1 \, d\varphi_2\cdots d\varphi_{N-1}.$$
Note 2: Your $f$ is undefined for $\alpha<0$ at $x=0$. To resolve this problem you can set $f$ to be $$f(x)=\begin{cases} |x|^{\alpha},&\mathrm{if}~x\neq 0, \\ 0 ,&\mathrm{if}~x=0, \end{cases}$$ and use the same proof.