Hello and new hoe.
Since I forgot to record for the.
Synchronoss lecture about this example.
I have to do it now.
So example.
The function F 04 Y.
Is presented algebraically, here and on the picture we have the red graph, so function equals.
04 negative X equals so 14X bigger than the 1 = 1/2.
In the middle part.
Equals 3/4 here equals 1/4 here. And so on and so forth.
Compare with the counter said.
Clearly this function is increasing.
Well, to say better, not decreasing.
So one can introduce the measure on intervals. So for any interval AB or we define the measure as F of B -- F of a.
So of course my this merger is far different from the Lib merger.
Nevertheless, one can define now sets out a measure.
And just to extend?
By this definition, from intervals to arbitrary Borel sets.
Eventually obtain measure on the Borel subsets of the interval.
And we don't need any further Sigma fields. In fact it can be extended.
Two really big measurable Sigma field and even further.
So in this example, I would like to compute this integral. Is there other technical exercise? But it would be good if you understand it.
So every one interval of the growth.
Or function F.
Here's our this.
Form.
Look when N = 0 function increases from zero.
21
so when N = 1.
We see that the function is constant in the middle part and it increases on the interval.
Zero up to 1/3.
And from 2 third from .2 / 3 up to one.
So it increases on such intervals.
When N = 2, we have four intervals where function increases.
From zero to 1 ninth.
From 2 ninth.
Up to 3 nines.
From 6 / 9 up to 7 / 9 and from.
8.
Over 9 up to 9 / 9 so this A or the numbers of intervals where function increases in case N = 2 and so on.
If any cause three, we have 8.
Intervals over function increases.
Remember it on everyone. Step on the middle part.
Function is constant.
Here, there and everywhere.
So we approximate function F.
Or for X = X from below by constant values on all those intervals. So here is the in this.
Scraf
function flight 2.
Corresponding to end being 2.
Is just collection of.
Intervals.
You will. You did this exercise in assignment.
One or two.
Approximation of function.
F of X = 6.
On the interval of the 11th 1 / 9 so I consider the case.
N = 2 so 1 / 9.
2 /, 9 and so on. This is function 5 two.
Simple function.
And.
The sequence of such functions so will approximate how they function F.
OK.
Define the integral according from the 1st.
From the front using the definition.
Consider integrals for simple functions and.
Try to understand what is the limit.
For a fixed an or the integral over the corresponding.
Particular interval is.
I / 3 two and multiplied by 1 /, 2 three and so again.
I tried to clarify this.
Using.
Oh, the interval.
2 / 9 up to 3 / 9.
The increment of function.
The value of this simple function is 1/4 of the increment the the the value of the function.
Is 2 / 9.
And the increment of the function capital F is 1/4.
So.
Here is bigger picture or the value of the simple function is 2 / 9.
And the increment of the area and function.
Equals 1/4, so this is.
Good response too.
For this interval.
The value of simple function to an ion multiply by the increment of the read function capital F.
Uh, this is explanation of this formula. Of course, on the another interval of the increment of the read function is the same, but the value of the simple function of will be different.
And do it take into account only the intervals over a red function increases.
So are the kids step here when N increases to N + 1?
Everyone interval of importance is split into 3 sub intervals, but the middle one has measure 0.
Well, look at this initial interval one.
23 Not over nine was split into three parts, and the read function is constant in the middle.
So the middle part has measure 0, so after we split.
On this interval.
Don't Lions 39th in three parts?
We have the following picture.
But the red function takes value 3 / 8.
In the mid part.
And it increases.
Bye.
1 / 8
On the 1st part and on the last part.
The black function simple function, of course again changes, so it is the same.
2 /, 9 On the 1st interval, the middle part is of no importance becauses the increment of red function is 0 the value of the measure is 0.
And the value on the last part is.
What 8 / 27?
So now we have the previous value to over 9.
Damn 7 or 8 + 8 / 27.
Times or 1/8.
So are the two subintervals.
But his ult.
In this formula and in that formula.
So here are two corresponding expressions.
Illustrating of this lion.
In the election notes.
So after we calculate, we have exactly the same value.
2 / 9 times quarter.
And the increment two over.
6 ^3
so this is illustration to this sentence in the lecture notes.
The increment appears as many times as there were intervals of the growth of the function.
So in this case.
We had.
For intervals.
When N = 2.
And.
So this increment.
Must be multiplied.
By 4
So in all the illustrating pictures, an or was equal to two.
And the end of the reasoning is just algebraic, purely algebraic calculation.
So if this simple function has number N.
For the value of integral is XN.
The initial value is 0.
Becausw if we have only one interval.
Only one interval on the picture.
Are they approximating?
Simple functions.
Use of course 0.
This is 50.
Because we consider just the whole interval 01.
And as it was shown, every next time when N increases the value.
Well, this integral increases by this number.
RGC also 1/3 to the power N + 1.
So a limit of XN.
Is does the.
Total sum of geometrical series.
From zero to Infinity, starting from 1/3. So we have 1/3 multiplied by the stranded geometrical series.
And the result is 1/2.
Everyone must understand.
Geometrical serious.
For those who do not remember, I will remind.
Some.
Off or or to buy.
From zero to Infinity equals 1 / 1 -- 0.
Of course are or must be smaller than one.
So you know a case where we have.
41 divided by.
2 / 3 choose 3 / 2.
And.
After we multiply by 1/3.
But the result is 1/2.
And the very last line.
Well, the sequence of simple functions.
Increases to earth.
Point twice.
And we can use the monotone convergence theorem. Or you can refer to the dominated convergence theorem.
So integral X, which is the limiting function.
Coincides with the limit of integrals of simple functions which approximate function F, so it is a limit of XN.
For rich equals half.
As it was shown.
2 minutes ago.
Thank you for attention and good night.