measure theory - If$mu$ is $sigma$- finite, $epsilon>0$, there exists $Ain mathcal{A}$ such that $mu(A)int f$

Problem

Let $X\mathcal{A},\mu$ be a $\sigma$-finite measure space. Suppose $f$ is non-negative and integrable. Prove that if $\epsilon>0$, there exists $A\in \mathcal{A}$ such that $\mu(A)<\infty$ and $$\epsilon+\int_A f \ d\mu>\int f \ d\mu.$$

Attempt

First, I will prove two propositions.

Proposition 1: If $f$ is non-negative measurable function, then $$\lim_{n\to \infty} \int \min(f,n) \rightarrow \int f$$

Proof: Note that $\min(f,n)$ increases to $f$ so the result follows from the monotone convergence theorem.

Proposition 2: Suppose $\mu$ is $\sigma$-finite and $f$ is integrable. Then given $\epsilon>$, there exists $\delta$ such that $$\int_A |f(x)| \ d\mu < \epsilon,$$ whenever $\mu(A)<\delta$.

Proof: Given $\epsilon>0$, from proposition $1$ we may find $n_0$ such that $\int_A |f(x)| \ d\mu - \int_A \min(|f(x)|,n_0) \ d\mu<\frac{\epsilon}{2}.$ Take $\delta<\frac{\epsilon}{2n_0}$. then:
$$\begin{align*} \int_A |f(x)|&=\int_A |f(x)| -\min(|f(x)|,n_0)\ d\mu+\int_A \min(|f(x)|,n_0)\ d\mu \\ &< \frac{\epsilon}{2}+n\mu(A) \\ &<\epsilon, \end{align*}$$
as soon as $\mu(A)<\delta$.

Finally we prove the result. $\epsilon>0$, use proposition $2$ to insure that $\int_B f\ d\mu <\epsilon$ whenever $\mu(B)<\delta$. Let $B^c=A$. Then
$$\begin{align*} \int f \ f\mu&=\int_A f\ d\mu+\int_B f\ d\mu \\ &<\int_A f \ f\mu+\epsilon, \end{align*}$$

as desired.

Concerns

For proposition $2$ as well as for the problem, I did not use $\sigma$-finiteness. I cannot see anything wrong with proposition $2$. However, in my solution to the problem, I have not guaranteed that such sets (i.e $B$ such that $\mu(B)<\delta$) actually exist. Please let me know what I did wrong here.