I want to check if my solution is right. Can someone help me?
Consider the measure space $(\mathbb{R}, 2^{\mathbb{R}}, µ)$ where $µ$ is the counting measure, hence the measure of a set is its cardinality. Show that if $f : \mathbb{R} → \mathbb{R}$ is $µ$-integrable then $E = \left \{x ∈ \mathbb{R}|f(x) \neq 0\right \}$ is finite or countably infinite.
So we have $\int_{E}fdµ=\lim_{n \rightarrow \infty}\int_{\phi_N \text{ simple, }\phi_n \leq f}\phi_n d µ=\sum_{i=0}^{\infty}a_i µ(A_i)$
If $E$ is infinite than we would have that the sum is infinite and it would not follow integrability.
Is my statement right?
Contrapositive approach:
If $E$ is not countable then for positive integer $n$ large enough the set: $$E_n=\left\{x\in\mathbb R: |f(x)|>\frac1n\right\}$$ is not countable.
This fact enables us to find distinct elements $x_i$ with $|f(x_i)|>\frac1n$ for $i=1,2,\dots$.
Then: $$\int|f(x)|d\mu\geq\sum_{i=1}^{\infty}|f(x_i)|=\infty$$ showing that $f$ is not $\mu$-integrable.