# If $f(x)$ is discontinuous at $x=0$, can $\int_{-1}^1 f(x)dx$ exist.

I am interested in the reasoning.

All help is appreciated

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

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.

• the plural of infimum is infima. For supremum is suprema. – Ittay Weiss Jul 19 '14 at 3:16

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 $0$.

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

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?