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?

  • 2
    $\begingroup$ Im confused by your proof. You've shown the infimum is less than or equal to the supremum. Yet to be Lebesgue integrable, the two should be equal. $\endgroup$
    – Alex R.
    Commented May 12, 2017 at 20:01
  • 1
    $\begingroup$ @AlexR. They are equal when $f$ is Riemann integrable since $\inf \{\int_a^b\phi : \phi \le f, \phi \text{ step function} = \sup \{\int_a^b\phi dx : \phi \ge f, \phi \text{ step function} \}$. $\endgroup$
    – SAS
    Commented May 15, 2017 at 16:38
  • $\begingroup$ Just nitpicking, but the second identity holds only when $f$ is a bounded function. This is because the right-hand side is not always equal to the Lebesgue integral. $\endgroup$ Commented May 16, 2017 at 5:10
  • 1
    $\begingroup$ Function is assumed bounded here. $\endgroup$
    – SAS
    Commented May 16, 2017 at 21:50

2 Answers 2


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.


Your idea is good but the right chain of inequalities is the following \begin{align} &\sup \left\{\int_a^b\phi : \phi \le f,\quad \phi \text{ step function} \right\} \\ =\,&\sup \left\{\int_{[a,b]}\phi : \phi \le f,\quad \phi \text{ step function} \right\} \\ \le\, &\sup \left\{\int_{[a,b]}\phi : \phi \le f,\quad \phi \text{ simple function} \right\} \\ \le\, &\inf \;\left\{\int_{[a,b]}\phi : \phi \ge f,\quad \phi \text{ simple function} \right\} \\ \le\, &\inf \left\{\int_{[a,b]}\phi : \phi \ge f,\quad \phi \text{ step function } \right\} \\ =\, &\inf \left\{\int_a^b\phi : \phi \ge f,\quad \phi \text{ step function } \right\}\end{align}


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