The statement is true because one can show there exists $\xi \in (a,b)$ such that $$\int_a^b f(x) \, dx \geq f(\xi) (b-a)$$
In fact assume that $$f(x) > \frac 1{b-a} \int_a^b f(y)\,dy$$ for all $x \in (a,b)$
Changing the values of $f$ at $a$ and $b$ if necessary (which doesn't alter the value of the integral), we have a new function $\hat{f}$ such that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy$$ for all $x \in [a,b]$
Remember now the theorem discussed in this question.
Since $\hat{f}$ is continuous at some $c \in (a,b)$, we can find $\varepsilon > 0\,$ so that $$\hat{f}(x) > \frac 1{b-a} \int_a^b \hat{f}(y)\,dy + \varepsilon$$ for all $x \in (c-\varepsilon,c+ \varepsilon) \subset (a,b)$.
Then, if we consider the partition $P=\{a,c-\varepsilon,c+\varepsilon,b\}$, we obtain $$\int_a^b \hat{f}(x)\,dx \geq L(\hat{f},P) \geq \int_a^b \hat{f}(x)\,dx + 2\varepsilon^2 > \int_a^b \hat{f}(x)\,dx$$ which is absurd.
The paper Rodrigo Lopez Pouso, Mean Value Integral Inequalities , Real Anal. Exchange Volume 37, Number 2, (2011), 439-450 is worth reading not only as the source of this proof.
Your proof is valid for a bounded function defined on the closed, bounded interval $[a,b]$, despite the apparent simplicity. It also relies on the fact that the Riemann and Lebesgue integrals are the same for step functions, as mentioned by Tony Piccolo.
The other proof you mention is also valid but takes you on a more roundabout path because it strings together a number of results, each of which is not altogether trivial to prove.
A bounded, measurable function defined on a set of finite
measure is Lebesgue integrable.
If a sequence of measurable functions converges almost everywhere
to $f$, then the limit function $f$ is measurable.
If $f$ is Riemann integrable on $[a,b]$, then there exists a
sequence of simple (measurable) functions converging almost everywhere
to $f$.
Adding even more complexity, the proof of the third statement that I know also uses the fact that the set of discontinuities of a Riemann integrable function must be of measure zero. It begins by constructing a partition of $[a,b]$ with dyadic intervals:
$$I_{n,k} = \begin{cases}[a + (b-a)\frac{k-1}{2^n}, a + (b-a)\frac{k}{2^n})\,\,\, k = 1 , \ldots, 2^n -1 \\ [a + (b-a)\frac{2^n-1}{2^n}, b] \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, k = 2^n\end{cases} $$
With $m_k = \inf_{I_{n,k}}f(x)$, we can construct the sequence of simple functions
$$\phi_n(x) = \sum_{k=1}^{2^n} m_k \, \chi_{I_{n,k}}(x)$$
The sequence is increasing and, since $f$ is bounded, by monotonicity is convergent to some function $\phi$ such that $\phi_n(x) \uparrow \phi(x) \leqslant f(x).$ If $x \in I_{n,k}$ then the oscillation of $f$ over that interval satisfies
$$\sup_{u,v \in I_{n,k}}|f(u) - f(v)| \geqslant f(x) - \phi_n(x) \geqslant f(x) - \phi(x).$$
Thus, $f(x) \neq \phi(x)$ only at points where $f$ is not continuous, which must belong to a measure zero set if $f$ is Riemann integrable. Therefore, $f$ is almost everywhere the limit of a sequence of simple functions and is measurable.
I would agree that the second approach is "pretty complicated". However, I think it is not uncommon to find theorems with a variety of proofs ranging in complexity.
Best Answer
Let $\epsilon >0$. Then $f \leq \frac {f^{p}} {\epsilon ^{p-1}} +\epsilon$ because $f \leq \frac {f^{p}} {\epsilon ^{p-1}}$ whenever $f > \epsilon$. [The inequality is obvious when $f \leq \epsilon$]. Hence $\int f \leq \epsilon (b-a)$. Since $\epsilon$ is arbitarary we get $\int f=0$.