Problem concerning a necessary and sufficient condition for Lebesgue measurability

Problem concerning a necessary and sufficient condition for Lebesgue
measurability

I am seeking solution verification/feedback for the following problem.
Thank you very much in advance.
$\textbf{Problem:}$ Let the function $f: [0,1] \rightarrow \mathbb{R}$ be
a non-negative, Lebesgue measurable function, and let $\lambda$ denote the
Lebesgue measure on $[0,1]$. Prove that $f$ is Lebesgue measurable if and
only if $\sum^{\infty}_{n=1} \lambda(\{x: f(x) \geq n\})$ converges.
$\textbf{Solution:}$ Assume $f$ is Lebesgue measurable. Then, $\int f \>
d\lambda$ is finite. Put $E_n=\{x: f(x) \geq n\}$. We show
$\sum^{\infty}_{n=1} \lambda(E_n)$ converges. For each $n$, $n\chi_{E_n}
\leq f$. Hence, $$ n \int \chi_{E_n} \> d\lambda \leq \int f \> d\lambda$$
$$ n \lambda(E_n) \leq \int f \> d\lambda$$ $$\lambda(E_n) \leq
\frac{1}{n} \int f \> d\lambda$$ Now, we see that $\lambda(E_n)
\rightarrow 0$ as $n \rightarrow \infty$; as such, we conclude
$\sum^{\infty}_{n=1} \lambda(E_n)$ converges.
$\>\>\>$Conversely, again put $E_n=\{x: f(x) \geq n\}$ and assume
$\sum^{\infty}_{n=1} \lambda(E_n)$ converges. Define a sequence of
functions $(\phi_n)$ by the following equation: $$\phi_n = \sum^{n}_{k=1}
\chi_{E_k}.$$ This sequence is monotone and converges to $f$. The Monotone
Convergence Theorem gives us that $$\int f \> d\lambda = \lim \int \phi_n
\> d\lambda.$$ Since, $$\lim \int \phi_n \> d\lambda = \lim_{n}
\sum^{n}_{k=1} \lambda(E_k) = \sum^{\infty}_{n=1} \lambda(E_n),$$ our
assumption gives us that $\lim \int \phi_n \> d\lambda$ and, consequently,
$\int f \> d\lambda$ is finite. $\blacksquare$