# How to prove that a function is Riemann integrable if and only if it is Darboux integrable?

(Pg. 16 Exercise 1.1.22. in Terry Tao’s Measure Theory) Show that if $f:[a,b]\to\mathbb{R}$ is a bounded function, then it is Riemann integrable if and only if it is Darboux integrable such that its Darboux and Riemann integrals are equivalent. In other words $$\int_a^b f(x) \ dx=\lim_{\|P\|\to 0}\mathcal{R}(f,\mathcal{P})\Longleftrightarrow\sup_{g\le f;\kern 0.1em\mathrm{piecewise \kern 0.1em constant}}\textrm{p.c.}\int_a^b g(x) \ dx =\inf_{h\ge f,\kern 0.1em\mathrm{piecewise \kern 0.1em constant}}\textrm{p.c.}\int_a^b h(x) \ dx.$$ Could anyone give me any hints?

For any fixed function $f$ on $[a,b]$ and for any tagged partition $\dot{\mathcal{P}}$ of $[a,b]$, we have that $$L(f,\mathcal{P})\leq \mathcal{R}(f,\dot{\mathcal{P}})\leq U(f,\mathcal{P})$$ which implies that Darboux integrable implies Riemann integrable, as noticed by herb steinberg.

Conversely, if $f$ is a bounded function on $[a,b]$ and $\mathcal{P}$ is any partition of $[a,b]$, then for any $\varepsilon>0$ we can find two taggings of $\mathcal{P}$, let's say $\dot{\mathcal{P}}$ and $\ddot{\mathcal{P}}$ such that $$\mathcal{R}(f,\dot{\mathcal{P}})\leq L(f,\mathcal{P})+\varepsilon\quad\text{and}\quad U(f,\mathcal{P})-\varepsilon\leq \mathcal{R}(f,\ddot{\mathcal{P}})\,.$$ This implies that a bounded, Riemann integrable function is Darboux integrable.

Now we just need to show that a Riemann integrable function is indeed bounded. The problem with a function which is not bounded is.... it's not bounded! This means that no matter how small the mesh of any partition of $[a,b]$ is, there will be some subdivision in there on which $f$ is not bounded. This fact can quite dramatically counteract the smallness of any mesh.

More explicitly, suppose that $f$ is Riemann integrable and unbounded. (Let me assume that $f$ is unbounded from above; the argument can be manipulated for the other case or applied to $-f$.) Let $L$ denote its Riemann integral. Let $M>0$ be any (large) number whatsoever. Let $\varepsilon>0$ be arbitrary. Let $\delta>0$ be chosen such that whenever $\dot{\mathcal{P}}$ is any tagged partition of $[a,b]$ with mesh smaller than $\delta$ we can say that $$|\mathcal{R}(f,\dot{\mathcal{P}})-L|<\varepsilon$$ Now let $\delta_0$ be any (tiny) number whatsoever such that $0<\delta_0<\delta$. Let $$\dot{\mathcal{P}}=\{a=x_0\leq t_0 \leq x_1\leq t_1\leq x_2\leq\cdots\leq x_{n-1}\leq t_{n-1}\leq x_n=b\}$$ be a tagged partition of $[a,b]$ with mesh smaller than $\delta_0$. Since $f$ is unbounded on $[a,b]$, there is some subdivision, let's say $[x_k, x_{k+1}]$, of $\mathcal{P}$ on which $f$ is unbounded. In particular, we can choose a (possibly new tag) $t^*$ from $[x_k, x_{k+1}]$ so that $$f(t^*)(x_{k+1}-x_k)>L+M-\sum_{j=0, j\neq k}^{n-1}f(t_j)(x_{j+1}-x_j)\,.$$

Let $\ddot{\mathcal{P}}$ denote this new partition with $t^*$. We then have $$L+M<f(t^*)(x_{k+1}-x_k)+\sum_{j=0, j\neq k}^{n-1}f(t_j)(x_{j+1}-x_j)=\mathcal{R}(f,\ddot{\mathcal{P}})< L+\varepsilon\,.$$ This is clearly nonsense. Thus, $f$ must be bounded.

• I think you have missed the important link about limit of Darboux sums as $|P|\to 0$ and the supremum/infimum of Darboux sums. That these are equal is non-trivial. – Paramanand Singh Feb 5 '18 at 12:08
• I treated these somewhat cursorily and just enough to show why they're intuitively correct. They should be made more formal, but these are the inequalities and ideas that one could use in the proofs. The OP seemed to just need the boundedness issue resolved. – Robert Wolfe Feb 5 '18 at 14:15

I can provide a brief outline and you can develop that into a proper proof. First you need to understand that the definition of Darboux integral as given in your post is equivalent to the formulation given in most textbooks. Thus given a bounded function $f:[a, b] \to\mathbb {R}$ we take an arbitrary partition $$P=\{a=x_0,x_1,x_2,\dots,x_n=b\}, x_{i-1}<x_{i},i=1,2,\dots,n$$ of $[a, b]$ and define Darboux sums $$L(f, P) =\sum_{i=1}^{n}m_{i}(x_i-x_{i-1}),\, U(f,P)=\sum_{i=1}^{n}M_{i}(x_i-x_{i-1})$$ where $$M_i=\sup_{x\in[x_{i-1},x_i]}f(x),\, m_i=\inf_{x\in[x_{i-1},x_i]}f(x)$$ Let $\mathcal{P}[a, b]$ denote the set of all possible partitions of interval $[a, b]$. The Darboux sums are themselves bounded ($L(f, P) \leq M(b-a), U(f, P) \geq m(b-a)$ where $M, m$ are supremum and infimum of $f$ on $[a, b]$) and thus $$\overline{J} =\inf_{P\in\mathcal{P} [a, b]} U(f, P), \, \underline{J} =\sup_{P\in\mathcal{P} [a, b]} L(f, P)$$ exist. You need to show that $$\underline{J} =\sup_{g\leq f, \text{ piecewise constant}}\text{p.c.}\int_{a}^{b}g(x)\,dx,\,\overline{J}=\inf_{h\geq f, \text{ piecewise constant}} \text{p.c.}\int_{a}^{b}h(x)\,dx$$ which establishes the equivalence of definitions of Darboux integral as given by Tao and as given in other textbooks.

Once this is done you need to show that $$\underline{J} =\lim_{|P|\to 0}L(f,P),\,\overline{J}=\lim_{|P|\to 0}U(f,P)$$ This is already done in this answer.

Since a Riemann sum is always sandwiched between upper and lower Darboux sums, it follows that if $f$ is Darboux integrable then both upper and lower Darboux sums tend to a common limit as $|P|\to 0$ and therefore so do the Riemann sums and the value of the integral defined by these approaches is also same.

To go the reverse way (from Riemann to Darboux) one just needs to prove that one can find a Riemann sum as near to a Darboux sum as we please by choosing the tag points $t_i\in[x_{i-1},x_i]$ such that $f(t_i)$ is near $M_i$ (or $m_i$ as needed). And thus if Riemann sums tend to a given value then the Darboux sums also do the same.

The approximation for a Riemann integral is bounded above and below respectively by the Darboux sup and inf. Therefore Darboux implies Riemann.

Riemann sums can be constructed as Darboux upper and lower sums. Therefore Riemann implies Darboux.

• How do you know that is is bounded above and below? – user522521 Feb 4 '18 at 21:52
• @majormaki If it isn't bounded, you can easily show that the filter of Riemann sums don't converge to anything with an appropriate choice of tag... – Robert Wolfe Feb 5 '18 at 4:01