Good evening so integrable functions.
As usual, we have an Lib measurable subset.
And.
The measurable function such that.
Positive part and negative part are both.
Find it, go find it integrals.
So on the picture of you have.
Function of which may be positive and negative. F Plus is the positive part.
Like this?
F minus.
Is the negative part.
And to be more accurate.
One has to write it seem in symmetrical way like this.
This is F minus and this is another.
Part of F minus.
So this is standard notations of plus F minus.
And in case of this tool.
But you could have parts of the function Fr integrable.
Here for these two integrals are finite.
But then it is natural to call their difference as the integral of function F.
So so far and in the future we consider only integrable functions, meaning that they both.
Parts.
Positive and negative are fine it.
So for students who are interested in something else.
If.
Are they first bought?
Is Infinity.
And.
This second part.
Is smaller than Infinity. Both integrals are not negative.
Or vice versa.
Well then.
Air function earth is called.
Quasi integrable?
That is just for your general knowledge, we will not consider such functions in this module, but for quasi integrable functions.
Either this term is Infinity and this is fine, it the result will be plus Infinity or on this part is fine. It and to this is Infinity. The result will be minus Infinity, so integrals are also well defined for quasi integrable functions.
But we will not consider such things in this module.
We say that function is integrable if everything is fine it.
The set of all functions which are integrable over a fixed domain here is denoted in this way. This is capital L.
L1
properties of integrals.
They are intuitively clear. Of course they can be rigorously proved, but it is enough if you understand and remember for these properties.
If one function is smaller than another of their integrals, again satisfy the same inequality.
So if functions F&G are integrable, the sum is also integrable and.
We have this.
Formula.
A function F is integrable and C is a number.
Well, then integral of CF Becausw ship times integral air.
So this properties number two and three.
Mean that the space of integrable functions is vector space.
If you studied.
So the module select two for one of them.
But you will understand this.
Next
if this inequality holds for all subsets.
From M for all lebec measurable subsets, then for this inequality holds almost everywhere.
So in particular, if we have this equality for all subsets or then function F = G Almost everywhere.
Integrable function is almost everywhere, fine it so we.
It can take the value plus or minus Infinity, but at particular points, so it must be finite almost everywhere.
But we usually consider only finite valued functions without Infinity. But remember Infinity is not a number.
For integrable function F.
So this obvious inequalities or take in from Earth.
So premium F multiply by the measure of the set A.
And to this.
Inequalities are obvious.
#7 is again obvious and very important.
Absolute value of the integral is smaller or equal then.
Integral of absolute value.
Or function.
Your function is not negative, integral zero or then function if F is zero almost everywhere.
If if it was J almost everywhere and F is integrable or then so is gene But the integrals coincide.
And #10.
Ritual bit generalized.
Very soon.
If.
Two sets are disjoint.
And function F is integrable separately on A&B, but then F is integrable on the Union and the integral of over the Union equals sum of the integrals.
This theorem can be generalized in the following way.
You function F is not negative or then the mapping from the sets.
So the integrals.
Do real numbers.
This mapping is a measure.
But the proof is not difficult. Again, we need simply to check the.
Definition The axioms of measure.
So we denote mu of a or the value of this integral, so it is not negative.
So this mapping has not negative values and so suppose we have sequence of.
We're wise disjoint measurable subsets.
Or for real number of real numbers.
We're talking about integrals over real numbers.
For the sequence of functions, GM F. Of X times in Decatur.
Increases.
Remember what indicator is.
If we have.
One said you want it too and we multiply function F by the indicator of the Union of the health red graph, which is 0 outside you want outside it too.
If we add one additional.
Set.
But then.
But the resulting function revealed.
Increase.
It remains is 0 outside and becoms.
Somewhat positive on the additional set you 3.
So this sequence of functions increases.
And converges to the function F of X time.
Is.
Interval function over the total union.
Now.
Or is she of that mu of the Union? OK, by definition is integral of G.
And remember, G is the.
Limiting function.
We know the monotone convergence theorem.
Functions JN converge to G monotonically increasing, so the limit of integral equals integral of the limit. It is illegal to swap integral and limit.
Finally.
For every GN.
We have
integral over the Union I.
And to this is sum of particular integrals becausw the submission is fine, it so this is fine it some of mu I.
And.
Using.
Using this.
Property.
We have
for the final.
Formula limit of or the sum equals integral of G equals measure of the Union.
So this is the basic property axiom of a measure, and the theorem is proved.
So remember that mu of EI?
Can equal Infinity.
A couple of examples.
If we have function.
Which is indicator simplest error function.
Well then mu of a.
Integral F / A.
Use the following measure if a subset of 020 there it is just leibig measure.
If a subset of.
The complement 02. Then the measure value of measures 0.
And for an arbitrary said, which is Libyan measurable?
Mu of a.
Integral of this simple function equals measure of the intersection.
The next example is a bit more interesting.
We have
pawsitive.
But not negative function such that the total integral is.
1.
For example, you can take.
So the density of the standard normal distribution.
So the total value is 1.
But the mapping from subsets to integrals is a measure, so in this particular case.
It is probability measure.
And in this very special case, it is probability measure corresponding to the normal distribution. So for any set a.
But the value of the measure is.
For the integral.
Under the curve.
So mute off.
He is integral over a.
If the M.
For this set function is a measure as it was proved by recently the total value is 1.
And.
It will have particular function F.
The value of the integral equals the probability.
Of the event.
Zed belongs to a, so that is standard normal.
Random variable.
That belongs to a is event.
And the value of the integral is exactly the probability of such event.
Thank you for your patience.
Next time we will consider very important theorem about dominated convergence.
Thank you for attention.