Suppose $f$ is integrable over $E$. And the assumptions are as given above. Then, currently, I have from Chebyshev's inequality,
$$ \lambda( \{ f > n \} ) \leq \frac{1}{n}\int_E |f| $$
Thus, since $f$ is integrable over $E$, we know $\int_E |f| < M$ for some $M \in \mathbb{R}$. Taking the limit of both sides yields,
$$ \lim_{n \rightarrow \infty} \lambda(\{f > n \}) = 0 $$
So it appears that I get the following result,
$$ \lim_{n \rightarrow \infty} n \cdot \lambda(\{f>n\}) = \infty \cdot 0 $$
So I feel I need to control how $\lambda(\{f >n \})$ is going to zero. But all I know is that it is going to zero linearly. Am I missing something?
WLOG assume $f \geq 0$ or otherwise just replace $f$ by $f^{+}$. Note that $E_n = \{f > n\}$ is a descending sequence of sets. Consider $f_n = f \chi_{E_n} \rightarrow 0$ a.e. and $f_n \leq f$. Using LDCT, we have
$$ 0 = \lim_{n \rightarrow \infty} \int f_n \, \text{d}\lambda \geq \lim_{n \rightarrow \infty} \int n \chi_{E_n} \, \text{d}\lambda = \lim_{n \rightarrow \infty} n \lambda(E_n)$$