00:01
Hide it in this problem which is given that eb belongs to r and e is smaller than b, gx and fx are two functions that are given and we are required to prove that f plus d of x is not integrable using dabox theorem and integrability epsilon formulation theorem.
00:15
So let us start with the solution.
00:18
In the first part we have to use the darbox theorem to prove that f plus d of x is not integral.
00:30
So firstly we need to find the f plus d of x.
00:33
Fx is given.
00:35
Equal to 1 when x belongs to rational numbers and negative 2 when x does not belong to rational numbers then we have g x defined as when x belongs to set of rational numbers and negative weight when x does not belong to set of racial numbers then the function f plus g of x will be defined as f plus g of x x is equal to when x belongs to rational numbers we have 1 plus 2 that is 3 so we write 3 when x belongs to set of rational numbers then we have when x does not belong to set of rational numbers negative 2 negative 8 it becomes negative 10 then x does not belongs to set of rational numbers so this was the function f plus 2 of x now let us suppose that a party of ab is done such that p is equal to ab such that p is equal to x not is equal to a x1 it is smaller than x2 smaller than equal to and so on up to x in minus 1 is smaller than x in is equal to b so this is the partition of ab now we will find the lower sum of f plus u of x for the partition p and it will be equal to summation of i is equal to 1 to n, m i xi, note that here mi represents the infirmum of mi is equal to infimum value of fx plus gx and x belongs to xi minus 1, xi minus 1, x so using these two values, the lower sum of the partition would be equal to infirmum value of f plus u of x is negative 10.
03:38
So we have negative 10 which is the constant.
03:42
We will take it out and in the summation we are left with summation i is equal to 1 to n dell of x i would be equal to x i minus x i minus 1 minus x i minus 1 and this would be equal to negative 10 x xx0 multiplied by now taking the values of i from 1 to n we will get x i minus x 0 plus x2 minus x1 plus and so on up to x n minus x n minus 1 x n minus 1 note that in this bracket we have minus 10 and in this bracket all the values would get cancelled except for x in n negative of x 0 so we have minus x n minus x 0 left with us and here minus 10 multiplied by this value x n is equal to b as shown in partition p and x0 is a so we have minus 10 multiplied by b minus a so let it be the equation one which gives the lower sum of f plus g of x for the partition p then lower sum of f plus g would be equal to the supreme value of lower sum of f plus u of x for the partition p x for the partition p that is equal to it will be equal to supreme of negative 10 multiplied by b minus a this will also be equal to minus 10 multiplied by b minus a this is the equation two now we will find the upper sum for the f f plus g for partition p, partition p, and it will be equal to summation of i is equal to 1 to n, mi, xi, where m i is the supremum value of the function fx plus g x and x and x is equal to x belongs to x i minus 1 to x i.
06:47
Now using these two values, we calculate this sum and it will be equal to summation of i is equal to 1 to n.
06:57
For mi we have 3, which is the supreme value of f plus t of x and for xi we have xi minus xi minus 1 and this will be equal to 3 multiplied by.
07:15
For xi and xi minus 1 we have the value as determined in part a, it will be equal to xi.
07:26
X i minus 1 and this whole summation can be calculated in a similar manner as in part a and it will be equal to b minus a therefore we write 3 multiplied by b minus a so this is the upper value of f plus g of x for the partition p now that we have the upper sum of f plus g for the partition p we can also find the upper sum of f plus g for the partition p we can also find the upper sum of f plus and it will be equal to infirm of upper sum of f plus g of x for the partition p and this will be equal to infimum value of 3 multiplied by b minus a this will be equal to 3 multiplied by b minus a now this is the equation 4 and we will use the equation 2 and 4 so from equation 2 and 4, we get that the lower sum of f plus d is not equal to the upper sum of the function f plus d...