We can say that a bounded function $f$ on a bounded interval $[a,b]$ is Riemann integrable if
$$\sup \{\int_a^b\phi : \phi \le f, \phi \text{ step function} \} = \inf \{\int_a^b\phi dx : \phi \ge f, \phi \text{ step function} \}$$
In the same way, I have seen the definition of Lebesgue integral as
$$\sup \{\int_a^b\phi : \phi \le f, \phi \text{ simple function} \} = \inf \{\int_a^b\phi : \phi \ge f, \phi \text{ simple function} \}$$
and the Lebesgue integral $\int_{[a,b]}f$ equals the lower Lebesgue integral on the left and the upper Lebesgue integral on the right.
SO we can prove that if the bounded function is Riemann integrable then it is also Lebesgue integral because every step function is also a simple function and
$$\inf \{\int_a^b\phi : \phi \le f, \phi \text{ step function} \} \le \inf \{\int_a^b\phi : \phi \ge f, \phi \text{ simple function} \} \\ \le \sup \{\int_a^b\phi : \phi \ge f, \phi \text{ simple function} \} \\ \le \sup \{\int_a^b\phi : \phi \ge f, \phi \text{ step function} \} $$
Yet in many places the theorem is proved by showing if $f$ is Riemann integrable then (1) there is an increasing sequence $\phi_n$ of simple functions bounded above by $f$ which converges; (2) the sequence converges almost everywhere to $f$; and (3) in this case $f$ must be measurable and Lebesgue integrable.
The second proof of (1), (2) and (3) is pretty complicated, yet the first proof seems so easy.
Am I missing something here? Is the first proof skipping some important step?
Best Answer
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.
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.