If a function $f(x)$ is Riemann integrable on $[a,b]$, is $f(x)$ bounded on $[a,b]$?

Question

Most statements regarding Riemann integrals (at least the ones that I have encountered) begin with the statement "for $f(x)$ bounded on $[a,b]$." I am wondering if Riemann integrability implies boundedness. I think that this has to be the case, but I am not sure. If Riemann integrability does imply boundedness, are improper integrals considered Riemann integrals? I would think that improper integrals wouldn't be Riemann integrals since improper integrals are allowed to be equal to $+\infty$ or $-\infty$. Or are improper integrals that are not equal to $\infty$ considered Riemann integrals?

I am confused.

Answer 1

Recall that a function is supposed to be (properly) Riemann integrable if for all $\epsilon>0$ there exists a partition such that $U-L<\epsilon$. Suppose $f$ becomes unbounded (say, unbounded above) near the point $x_0$. How are we to make sense of the upper sum $U$ when one of the intervals in the partition (the one containing $x_0$) has no supremum? The definition only makes sense when $f$ is bounded.

Nonetheless, if $f$ becomes unbounded near $x_0$, it still may be (improperly) Riemann integrable on an interval containing $x_0$. This is because we define the improper integral to be the limit of the proper integrals over the regions where we delete an open neighborhood of $x_0$, as we let that neighborhood get smaller and smaller.

If someone describes a function as "Riemann integrable" on a set $S$, it may be ambiguous if they mean properly or improperly (technically they should mean the former), and one has to infer from context whether they are implying that the function is bounded.

Answer 2

Yes, a Riemann-integral must be bounded; if not, then, for partition width $\|P\|$ , consider a point $x$ with $f(x)> \frac {M}{\|P\|} $ (which will exist for any $M$, since f is , by assumption, unbounded). Now, since $f$ is unbounded, $M \rightarrow \infty$ , so the Riemann sum associated to the partition length $\|P\|$ will be bounded below by $$\|P\|\left(\frac {M}{\|P\|}\right)=M \rightarrow \infty$$. Since this is true for any partition length $\|P\|$ then the Riemann sum will diverge.

Answer 3

There is a minor ambiguity in the term "Riemann integral": it tends to be used both for Riemann's original formulation -- which involves tagged partitions and requires convergence in a very strong sense: uniformly in the mesh (or norm) $||\mathcal{P}||$ of the partition $\mathcal{P}$ -- and also G. Darboux's later simplification in terms of upper and lower sums and upper and lower integrals, which is for most purposes technically easier to work with and thus is the one which is carefully developed in most undergraduate texts.

The ambiguity can be justified by the fact the Riemann and Darboux theories give different descriptions of what ultimately turns out to be the same linear functional: a function $f: [a,b] \rightarrow \mathbb{R}$ is Riemann integrable if and only if it is Darboux integrable (the hard part of this is to show that Darboux integrable functions are Riemann integrable) and when these conditions hold the associated real number $\int_a^b f$ is the same. For a careful exposition of the Darboux and Riemann integrals including a comparison between the two, see Chapter 8 of these notes.

I bring up the distinction between Darboux and Riemann because it is relevant to your question of boundedness of integral functions, and because of the two helpful answers already left to this question, one addresses the Darboux case and the other the Riemann case. Either way though the following simple observation lies at the heart of the matter.

For $f: [a,b] \rightarrow \mathbb{R}$, the following are equivalent:
(i) $f$ is bounded above (respectively, bounded below).
(ii) For any partition $\mathcal{P} = \{a= x_0 < x_1 < \ldots < x_{n-1} < x_n = b\}$, the restriction of $f$ to each subinterval $[x_i,x_{i+1}]$ is bounded above (respectively, bounded below).

So if $f$ is unbounded above, then for any partition $\mathcal{P}$, there is at least one subinterval $[x_i,x_{i+1}]$ on which $f$ is unbounded above, hence the upper sum $\mathcal{U}(f,\mathcal{P})$ does not exist as a real number, so we can't even define the Darboux integral. Alternately, if we want to work in the extended real numbers, we would say that if $f$ is unbounded above, $\mathcal{U}(f,\mathcal{P}) = \infty$. Similarly, if $f$ is unbounded below, $\mathcal{L}(f,\mathcal{P}) = -\infty$ (c.f. Proposition 8.2 in the linked notes). This means: if $f$ is unbounded above then $\overline{\int}_a^b f = \infty$, and if $f$ is unbounded below then $\underline{\int}_a^b f = -\infty$. With this extended definition we would define a function to be Darboux integrable if and only if its upper and lower integrals are both finite and are equal, so we see that Darboux integrable functions are bounded.

For the Riemann integral there is a similar argument: if $f$ is unbounded above, then no matter what partition we choose, then for any $M > 0$ there will be a tagging $\tau$ -- i.e., a choice of sample point $x_i^* \in [x_i,x_{i+1}]$ such that the Riemann sum $R(f,\mathcal{P},\tau) = \sum_{i=0}^{n-1} f(x_i^*)(x_{i+1}-x_i)$ is greater than $M$. This is a nice exercise: the idea is to use the above observation and choose one subinterval $[x_i,x_{i+1}]$ on which $f$ is unbounded above, choose the sample points in the other subintervals arbitrarily, and then choose $x_i^* \in [x_i,x_{i+1}]$ so that $f(x_i^*)$ is large enough to make the entire Riemann sum come out greater than $M$. Thus we see that if $f$ is unbounded above it cannot be Riemann integrable (but it's definitely a result, not a definition, in this case), and similarly if $f$ is unbounded below.

To address the rest of your question: yes, when one writes "Riemann integrable" one generally means to neglect the case of improper Riemann integrals, which of course can be finite even for some unbounded functions. This is true notwithstanding the fact that the same notation $\int_a^b f$ is used for Riemann integrals and improper Riemann integrals. Generally, anyway: in any particular case you should check to confirm that the terminology is being used in this way.

Added: The class of Riemann-Darboux integrable functions was characterized by Lebesgue (though my colleague Roy Smith has shown me a passage in Riemann's work showing that he had the result as well).

Theorem (Lebesgue Criterion) For a function $f: [a,b] \rightarrow \mathbb{R}$, the following are equivalent:
(i) $f$ is Riemann integrable.
(ii) $f$ is bounded, and the set of discontinuities of $f$ has measure zero.

Since this result was first published in the 20th century, one can correctly infer that it is not really needed in the development of the Riemann integral. From a pedagogical perspective, I would rather have students learn to use less heavy tools with more dexterity. Nevertheless I give a proof in $\S$ 8.5 of my notes which does not use any measure theory or even the theory of countable/uncountable sets. (The proof given there is not due to me; it is taken from notes of A.R. Schep.)

Note also that another answer to this question currently contains a false statement of this result. The characteristic function of the classical middle thirds Cantor set shows that "measure zero" cannot be replaced with "countable".

Answer 4

That an Riemann integrable $f:[a,b]\to\mathbb R$ is bounded can be proved directly from the definition of the integral via Riemann sums, without assuming the opposite and looking for a contradiction.

Recall that $f$ is Riemann integrable with an integral $I\in\mathbb R$ if

$$\forall\varepsilon>0\,\exists\delta>0\,\forall\tau\,\text{partition}:d\tau<\delta\to\forall\left\{ \xi_{i}\right\} _{i=1}^{n}:\left|R_{\tau,\xi}(f)-I\right|<\varepsilon,$$ where $d\tau$ is the diameter of $\tau$ and $R_{\tau,\xi}(f)$ is the Riemann sum $R_{\tau,\xi}(f)=\sum_{i=1}^{n}f(\xi_{i})\left(x_{i}-x_{i-1}\right)$.

Now by a "fix-take" procedure of values in and out of that predicate formula you can derive the boundedness of $f$:

• Fix your favorite $\varepsilon>0$, e.g. let $\varepsilon=1$.
• Take a $\delta>0$.
• Fix your favorite partition, e.g. $\tau=\left\{ a+i\frac{b-a}{n}\right\} _{i=0}^{n}$, where $n \in \mathbb N$ is large enough so that $d\tau=\frac{b-a}{n} < \delta$.
• Now we prove $f$ is bounded on any of the finitely many $[x_{j-1},x_j]$. As mentioned in other answers, this implies $f$ is bounded on the entire $[a,b]$.
  Fix such an interval by choosing $j\in\{1,\ldots,n\}$, and fix a set of sampling points $\left\{ \xi_{i}\right\} _{i\neq j}$ for the rest of $[a,b]$.
• The only unfixed variable now is the sampling point $\xi_j\in[x_{j-1},x_j]$. The formula says $$\forall\xi_{j}\in\left[x_{j-1},x_{j}\right]:\left|\sum_{i=1}^{n}f(\xi_{i})d\tau-I\right|<\varepsilon=1.$$ Divide by $d\tau$ and take $f(\xi_j)$ before the sum to see the result more clearly – for any $x\in\left[x_{j-1},x_{j}\right]$, $$\left|f(x)-\left(\sum\nolimits _{i\neq j}f(\xi_{i})+I/d\tau\right)\right|<1/d\tau.$$ But $\forall x \left|f(x)-a\right|<b$ mean exactly that $f$ is bounded, provided that $a,b$ are independent of $x$, as is the case here (everything's fixed).

I like this proof as it requires almost no thinking - you just unwrap step by step the predicate formula of the definition. Note that it is practically the same as what @Pete L. Clark suggested, no novelty here, but presented without the assumption-contradction part, which I prefer to avoid when possible. I also find this way easier to reproduce for exams :P