Hello and Nico.

So it is time to compare for the Libyan integral and Riemann integral.

Are you do this theorem please? Supposed to function from a finite interval is bounded.

Then

we have the following statements.

Function F is Riemann integrable if and only if it is almost everywhere continuous with respect to leibig measure.

And Secondly, if function F is Riemann integrable on the finite interval, then it is Libya integrable and the integrals are the same.

Remember that can easily happen or that a function which is bounded on the bounded interval is really big integrable but not Riemann integrable. There was exercise in one of the homework.

So this theorem is without proofs.

You are welcome to investigate all the textbooks.

And remember that any bounded measurable function.

On a finite interval.

Is Arabic integrable? So remember that everywhere here.

Or we consider.

The bounded interval AB.

Alright, what is the very human integral integral?

So this is of course a revision of your calculus, which is.

Absolutely important background.

So take the lower Riemann.

Some lower Riemann integral.

How it is constructed?

We divide the interval AB.

Into small.

Subintervals of the length of an overhead.

Here are the formula.

And after that.

Approximate the area under the function by rectangles.

So I'm talking about positive function.

At the moment.

Are they pictures presented? So we calculate all the areas of all rectangles and add together that is a riemann's construction.

What happens in case of Libya integration?

We divide the arrange.

Or function into small intervals.

And approximate the area under the function again by rectangles.

But the picture is like this.

The range of function F.

The range is on the vertical line is divided into small intervals, so this is very general.

Not very good's idea for you to better understand.

But what is a Riemann integration of? What is the Libya integration?

So we divide the arrange or function into small intervals and construct rectangles like previously, but the rectangles will be far different.

From the element case.

And again calculate the total sum of the rectangles. This is the approximations approximation from below to the Arabic integral. So very often the results are the same.

In case an approaches Infinity.

So in both cases.

In the limit we obtained the area under the curve.

So that in most cases the Riemann and LBCC integrals coincide.

So this is illustration to the.

Therium

bought.

2.

Your function is Riemann integrable, then it is LBCC integrable.

Although it was about bounded functions and bounded intervals.

Of course one can consider negative functions.

Now the integral will be negative, but the constructions are the same.

And the values of will go insight in case.

They are human integral exists.

It is interesting to look at the so called improper Riemann integrals.

So if the interval is not finite.

But the Riemann integral is defined as the limit.

When B approaches Infinity.

In case function is not negative.

And the improper integral Riemann integral exists may be finite or infinite.

But then they began to grow, exists and equals the improper Riemann integral. So this is.

A slight generalization of the previous theory about Infinity intervals.

But the same concerns.

Unbounded functions.

Or which for example approach Infinity?

And again, the improper Riemann integral is the limit.

When epsilon goes to 0.

And like previously, if the function is.

Not negative right then? Or the improper Riemann integral, if it exists, coincides with the.

Will it be again to grow?

Very important remark or the condition.

That function is positive is important. OK function can maybe of course just negative.

And everything remains correct for negative values of the Riemann integral. And really big.

But if function changes sign.

For instance, on the infinite interval.

Well then.

Something interesting can happen.

So consider this function which is on the screen.

On the infinite interval.

From zero to Infinity.

And try to compute the Riemann integral and Rebecca integral.

So the function looks like this.

We have interval from zero to Infinity.

And the function looks like follows on the interval.

01

It tickles plus one.

So minus 1 two 0 / 1.

On the next interval, the sign changes and the function equals.

Minus.

Half.

Have open intervals.

On the next interval, the function equals 1/3.

And it is positive.

On the next interval, it take also one quarter.

And it is negative.

And so on.

So this function is Riemann integrable.

According to the definition, calculation is.

Phone in the election notes.

Or we pass to the limit when N approaches Infinity. So on the 1st interval we have one, on the 2nd we have minus half plus 1/3 and so on.

This is.

Sam

Over the serious.

Which changes signs and the absolute value decreases.

From calculus of you know that search serious converges.

I don't remember the value, but it is just a finite number.

On the other hand.

But remember the definition of the alley Big Intagram we need to introduce.

Positive part of the function.

And try to compute the integral.

It will be plus Infinity.

And we must consider the negative part of the function. Try to calculate the Libya integral. It is again plus Infinity.

Why is it so becausw? So we have?

Integral.

Note Infinity.

F plus.

M What is it we take into account? Only positive?

Well, you saw the function.

We have one.

Velez

who won third.

Plus 1/5.

Plus one 7th.

And this serious diverges.

Similarly for the negative part of the function of the total value of the Libya integral is.

Infinite and according to definition of, except that this function is not integrally big integrable or the Libya green Toghrul is undefined.

So only in such cases.

We see that the Riemann integral exists and Rebecca Integral does not exist, so only in this case the Riemann integration is more general.

Alright, thank you for your patience next week. Where will.

Locate probability and see how the general in theory of Libya integration applies to.

Random variables and probability distributions. Thank you for your patience goodbye.