4
$\begingroup$

I am interested in the reasoning.

All help is appreciated

$\endgroup$
  • 6
    $\begingroup$ $$f(x) = \begin{cases} -1 &, x < 0 \\ 0 &, x = 0\\ 1 &, x > 0 \end{cases}$$ $\endgroup$ – Daniel Fischer Jul 18 '14 at 10:45
  • $\begingroup$ As far as I can (should) read the question is "I am interested in the reasoning. All help is appreciated". Can't understand it. $\endgroup$ – Git Gud Jul 18 '14 at 10:47
  • $\begingroup$ The change of the function value in a null set doesn't change the integral. $\endgroup$ – Shine Jul 18 '14 at 10:51
  • $\begingroup$ Piece-wise continuous functions are Riemann integrable. So all you need is to find out functions with discontinuity of first kind. $\endgroup$ – hrkrshnn Jul 18 '14 at 10:53
  • $\begingroup$ That doesn't seem to be correct @Shine: if you change the function in a countable set as to make it not-bounded then the integral doesn't even exist. $\endgroup$ – Timbuc Jul 18 '14 at 11:45
9
$\begingroup$

In general, when $f$ and $g$ differ in (at most) a finite amount of points (in the interval $[a,b]$), we always have $$ \int _a^bf(x)\,dx=\int_a^b g(x)\,dx $$ This follows from the definition using sums over (infima and suprema of) a division of the interval $[a,b]$ in smaller intervals and the fact that we can make arbitrarily small intervals around the points where the functions are different.

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ the plural of infimum is infima. For supremum is suprema. $\endgroup$ – Ittay Weiss Jul 19 '14 at 3:16
5
$\begingroup$

The integral is highly insensitive to small changes in the function. In particular, the function $f(x)=0$ for $x\ne 0$ and $f(0)=16$, as far as the integral is concerned, is the same as the constantly $0$ function. In particular, both are integrable and the integral is $0$.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Example: $f(x)=0$ for $x<0$ and $f(x)=1$ for $x>0$. Another example: $f(x)=\ln|x|$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Given the way you ask your question, I assume that you are unfamiliar with the definition of the Riemann Integral. Let the upper Riemann-sum $U$ of a function $f$ on a partition $P = \{x_0, x_1, x_2 \cdots x_n\}$ be defined as $$\sum_{i = 1}^n \Delta x_i\cdot I^*$$ where $\Delta x_i = x_i - x_{i-1}$ and $I^*$ is the greatest value of $f$ on $[x_{i-1}, x_i]$. The lower Riemann-sum $L(f, P)$ is defined in the obvious way.

Definition 1: A function is Riemann Integrable if there exists a partition $P$ so that for every $\epsilon > 0$ we have that $U(f, P) - L(f, P) < \epsilon$

Consider this slightly edited function provided by Daniel Fischer as a comment to your question: $$f(x) = \begin{cases} -1 &,-1 \le x < 0 \\ 0 &, x = 0\\ 1 &,1 \ge x > 0 \end{cases}$$

Let $P = \{-1, 0-\epsilon, 0+\epsilon, 1\}$. We have $U(f,P) = (-1) \cdot(1-\epsilon) + 1\cdot2\epsilon + 1\cdot (1-\epsilon) = -1 + \epsilon + 2\epsilon + 1 - \epsilon = 2\epsilon$ and $L(f,P) = (-1) \cdot (1-\epsilon) + (-1)\cdot 2\epsilon + 1\cdot(1-\epsilon) = -1 + \epsilon - 2\epsilon + 1 - \epsilon = -2\epsilon$. So we have that $U(f,P) - L(f,P) = 4\epsilon$. The function is still integrable, although our value is greater than $\epsilon$. Can you tell why? Can you see how you might configure the values of the partition so that the result is prettier?

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.