If the solution is too simple, it must be incorrect?

Question

Prove that to each $\epsilon > 0$ there exists a $\delta > 0$ such that $\displaystyle \int_E |f| d \mu < \epsilon$ whenever $\mu (E) < \delta$, where $f \in L^1(\mu)$.

I found this question asked on MSE here, here, and here, but some of the suggestions seem relatively complicated (at least compared to what I did). When I was solving the problem, no such ideas/theorems found in those links came to my mind; indeed, I just followed my nose by playing with inequalities. But, after looking at some other solutions, I am beginning to doubt that mine is correct. Here is what I did:

First note that by definition

\begin{align} \int_{E} |f|d \mu &= \sup \{\sum_{i=1}^m a_i \mu (A_i \cap E) \mid m \in \Bbb{N}, 0 \le a_i \le |f|, E = \bigsqcup_{i=1}^m A_i, A_i \mbox{measurable} \} \\ &\le \sup\{\max \{a_i\} \sum_{i=1}^m \mu(A_i \cap E) \mid ... \} \\ &= \sup\{\max \{a_i\} \mu(E) \mid ... \} \\ &= \mu(E) \cdot \sup\{\max \{a_i\} \mid ... \} \\ &\le \mu(E) \inf_{x \in E} |f(x)| \\ \end{align}

The first inequality follows simply because the elements in the one set are larger than the other. The second because $0 \le a_i \le |f|$ on $E$ implies $a_i \le \inf_{x \in E} |f(x)|$ and therefore $\max \{a_i \} \le \inf_{x \in E} |f(x)|$. Now, if $\inf_{x \in E} |f(x)| = 0$, then any $\delta > 0$ will do, since the above together with this implies $\int_{E} |f| d \mu =0$. If not, then $\displaystyle \delta = \frac{\epsilon}{\inf_{x \in X} |f(x)|}$ will work.

I can't spot the error.