Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be compact, define $$D^+(f):= \inf\left\{\int t:t\geq f, t= \text{step function}\right\}$$ $$D^-(f):= \sup\left\{\int t:t\leq f, t= \text{step function}\right\}$$ If $D_+(f)=D_-(f),$ we call $f$ Darboux integrable (equivalent to Riemann integrable) and the common value the Darboux integral of $f$.

How would I show that if $f$ is Darboux integrable, then $f$ is Lebesgue integrable, and $D^+(f)=D^-(f)=\int f$

  • 2
    $\begingroup$ See my edits to understand proper MathJax usage. $\qquad$ $\endgroup$ – Michael Hardy Apr 27 '16 at 20:37
  • $\begingroup$ There are functions which are Riemann integrable but not Lebesgue integrable, since if $f$ is Lebesgue integrable, then so if $\left|f\right|$ (and this is not true for Riemann integrable functions). Find such an example -- and here is your counterexample. $\endgroup$ – Guy Apr 27 '16 at 20:42
  • $\begingroup$ Well if you define the Lebesgue integral as $\int f =\sup\{\int h: h\leq f, \text{ and } h \text{ is simple}\}$, then clearly $D^-(f) \leq \int f$ since the set of integrals of step functions is a subset of the set of integrals of simple functions. $\endgroup$ – Trevor Norton Apr 27 '16 at 20:44

For a bounded interval consider WLOG $f :[a,b] \to \mathbb{R},$ a nonnegative Riemann integrable function. Then

$$\int_a^b f(x) \, dx = \sup_{\phi \leqslant f} \int_a^b\phi(x) \, dx = \inf_{\psi \geqslant f} \int_a^b\psi(x) \, dx ,$$

where $\phi$ and $\psi$ are step functions. This is straightforward to show using upper and lower Darboux sums and integrals.

Since any step function is a simple function, we have for simple functions $\hat{\phi}$ and $\hat{\psi}$

$$D^-(f) = \sup_{\phi \leqslant f} \int_a^b\phi(x) \, dx \leqslant \sup_{\hat{\phi} \leqslant f} \int_{[a,b]}\hat{\phi} \leqslant \inf_{\hat{\psi} \geqslant f} \int_{[a,b]}\hat{\psi} \leqslant \inf_{\psi \geqslant f} \int_a^b\psi(x) \, dx = D^+(f),$$

and Riemann integrability, $D^-(f) = D^+(f),$ implies Lebesgue integrability with

$$\sup_{\hat{\phi} \leqslant f} \int_{[a,b]}\hat{\phi} = \int_{[a,b]}f= \int_a^bf(x) \, dx .$$

The converse is not true. Lebesgue integrability on a bounded interval does not imply Riemann integrability. The Dirichlet function is a counterexample.


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