Find two non-Riemann integrable functions whose sum is not Riemann integrable.
An example of this is the function that is 1 on any rational number and 0 elsewhere.
And what would the second function be?
You can change 1 and 0.
The simplest example of non-integrable function are: 1/x in the interval [0, b]
To find two non-Riemann integrable functions whose sum is also not Riemann integrable, we need to understand the conditions for Riemann integrability and construct functions that violate these conditions.
In order for a function to be Riemann integrable, it needs to satisfy the following conditions:
1. Boundedness: The function must be bounded on a closed interval [a, b].
2. Almost Everywhere Continuity: The function can have a finite number of discontinuities, but they must be of the first kind (jump discontinuities) and have a zero measure.
3. Integrability of Discontinuity Points: The function should have only a finite number of discontinuities. In other words, the set of discontinuity points should be of measure zero.
To construct two non-Riemann integrable functions satisfying the question's requirements, let's consider the classic examples:
1. Dirichlet's Function (also called Thomae's function):
This function is defined as follows:
D(x) = 1, if x is rational
D(x) = 0, if x is irrational
The Dirichlet's function violates condition 2 because it is discontinuous at every point. Moreover, it is not Riemann integrable because its upper and lower Riemann sums do not converge to the same value.
2. Thomae's Function:
This function is defined as follows:
T(x) = 1/n, if x = m/n is rational (in lowest terms)
T(x) = 0, if x is irrational
Thomae's function satisfies condition 2 since it only contains removable discontinuities (which have a measure zero). However, it is not Riemann integrable since it has infinite points of discontinuity.
Now, let's consider the sum of these two non-Riemann integrable functions:
F(x) = D(x) + T(x)
The function F(x) will also violate the conditions for Riemann integrability. It will be discontinuous at every rational number, which means it will have infinitely many points of discontinuity. Therefore, the sum of Dirichlet's function and Thomae's function, F(x), is not Riemann integrable.