Let $c$ be inside $(a,b)$ and let $d$ be inside the set of all real numbers. Define $f:[a,b]\rightarrow R$ as $$f(x):=\cases{d\quad& \text{if $x$ is equal to }c\cr 0&\text{if $x$ is not equal to }c\cr}\ $$ Prove that $f$ is Riemann integrable and compute $\displaystyle \int \limits_{a}^{b}f$ using the definition of the integral.

To show it is Reimann integrable I know I have to show that $\sup L(p,f) =\inf U(p,f) $ (notation-wise this mean the $\sup$ (lower Darboux Sum) = $\inf$ (upper Darboux Sum)).

I am running into confusion determining these values though. I know $L(p,f)=0$ and I have $U(p,f)=d(b-a)$. Is this correct? If so how do I determine the $\sup$ and $\inf$?

  • $\begingroup$ The error is when stating that U(p,f)=d(b-a). This is incorrect, unless p is the trivial subdivision {a,b}. $\endgroup$ – Did Jul 18 '13 at 16:01
  • $\begingroup$ then what would U(p,f) be? I am running into confusion with determining it when I am working with an arbitrary number d. $\endgroup$ – user72195 Jul 18 '13 at 16:05
  • $\begingroup$ U(p,f)=dw(p) where w(p) denotes the length of the interval of the subdivision p containing c. (Unrelated: please use @ in comments.) $\endgroup$ – Did Jul 18 '13 at 17:16
  • $\begingroup$ @Did so then how would I go about determine what the inf of that is? I know it would have to be 0 because the sup of L(p,f) is 0 since the whole thing is 0? $\endgroup$ – user72195 Jul 18 '13 at 17:51
  • $\begingroup$ Use subdivisions p such that w(p) goes to zero. $\endgroup$ – Did Jul 18 '13 at 18:19

Hint: Details depend on precisely how the Riemann integral is presented, so this can only be a guide.

Take a partition $\Pi$ of the interval $[a,b]$. Then any Riemann sum $S$ based on $\Pi$ satisfies the inequality $$0\le S\le \epsilon d,$$ where $\epsilon$ is the mesh of the partition, that is, the maximum length of the subintervals of $\Pi$.

As $\epsilon\to 0$, $S\to 0$.


Some useful things to note:

This also works if a function disagrees with $0$ at finitely many points, just repeat Andre's argument $n$ times, surrounding each point of issue by an interval

$\frac{\epsilon}{nM}$ where $n$ is the number of points and $M$ is the max of these points (this may be done as it is finite list).

Notice that this stems from a more general issue that if $f \in \mathcal{R}$ and we define a set $ \{x_1, \ldots, x_k \}$, then we may define

$$g(x) : = \begin{cases} f(x) \qquad x \notin J \\ d_i \qquad x = x_i \end{cases}$$

Then we can prove that $g \in \mathcal{R}$ and even that $$\int_{[a,b]} f \, dx = \int_{[a,b]} g \, dx.$$

In fact, this even works if, say, $f$ and $g$ disagree on all of $\mathbb{Q} \cap [a,b]$!

  • $\begingroup$ Regarding your last sentence: But the Dirichlet function is not Riemann integrable! $\endgroup$ – philmcole Dec 30 '17 at 9:58

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