Does this proposition hold ?
I'm studying functional analysis and I come up with the situation where I (maybe) have to show this.
If $f:X\to \mathbb C$ is measurable simple function and $f\in L^p$, then $\mu(\{x\mid f(x)\neq 0\})<\infty$.
Here is my proof.
Since $f$ is simple, I can write $f(X)=\{a_1,\cdots, a_n\}$ where $a_i\in \mathbb C.$
Let $\alpha:=\max\{\tfrac{1}{|a_i|^p}\mid 1\leqq i\leqq n, a_i\neq 0\}.$
Then, I get $$\chi_{\{x\mid f(x)\neq 0\}}(x)\leqq \alpha|f(x)|^p \ \mathrm{for \ all}\ x,$$ since for each $x$, if $f(x)=0$ then this inequality is obvious and if $f(x)\neq 0$ then $0\neq f(x)=a_i$ for some $i$ and LHS $=1$, RHS $=\alpha |a_i|^p\geqq \tfrac{1}{|a_i|^p}|a_i|^p=1$.
Thus I get $$\mu(\{x\mid f(x)\neq 0\})=\int_X\chi_{\{x\mid f(x)\neq 0\}}(x)d\mu(x)\leqq \alpha\int_X|f(x)|^pd\mu(x) <\infty$$
Is this O.K.?
Yes your proof is good!
Another way to see this is to remember that f can be expressed as the finite sum of weighted characteristic functions supported over disjoint measurable sets.
So if we assume to the contrary that $\mu(\{f(x)\neq 0\}) =\infty$, then one of the characteristic functions in the expression of f must be supported by a set of infinite measure, but since this characteristic function will also be weighted by a non zero scalar, that would contradict that f is in $L_p$