Monotone convergence theorem

In mathematics, the 'monotone convergence Theorem' are some of the few theorems. Examples of major expansion here.

Convergence of a monotone sequence of real numbers

Theorem === === If <math> \ {a_n \} </ math> is a monotone sequence of real number s (for example, if a' _n ≤ a _''' n +1 or a''' _{n ≥ _a'} n +1 for every n ≥ 1), then this sequence a block limit if and only if the line side is survival. <ref> A generalization of this Theorem given by John Bibby (1974) "Axiomatisations the average and a further generalization of monotonic sequences," Glasgow mathematical Journal, vol. 15, pp. 63-65. </ Ref>

Proof === === We prove that an increase <math> sequence \ {a_n \} </ math> is bounded above, then it is convergent and is opposed <math> \ sup \ limits_n \ {a_n \} </ math>.

Since <math> \ {a_n \} </ math> is non-empty and the like, it is bounded above, then, from the at least Upper bound property by phone at real, <math> c = \ sup_n \ {a_n \} </ math> and government intervention. Now all <math> \ varepsilon> 0 </ math>, there exists <math> a_N </ math> such that <math> a_N> c - \ varepsilon </ math>, since otherwise < Math> c - \ varepsilon </ math> is a Upper bound of <math> \ {a_n \} </ math>, which contradicts the <math> c </ math> do <math> \ sup_n \ {a_n \} </ math>. Then since <math> \ {a_n \} </ math> is increased, <math> \ forall n> N, | c - a_n | = c - a_n \ leq c - a_N <\ varepsilon </ math>, hence by definition, the limit of <math> \ {a_n \} </ math> is <math> \ sup_n \ {a_n \}. </ Math>

Remark === === If a sequence of real numbers and reduced survival below, then infimum is forbidden.

Convergence of a monotone series

Theorem === === If all the natural numbers j and k,''' a _{j, k''} is a non- -not real number and a''' _{j, k' ≤'''} _{a j +1, k , then (see for example <ref> J Yeh (2006} } </ ref> page 168)}

<math> \ Lim_ {j \ to \ infty} \ sum_k Ã_ {j, k} = \ sum_k \ lim_ {j \ to \ infty} Ã_ {j, k}. </ Mathematics>

The Theorem that if you have an infinite matrix of non-negative real numbers such that

The line is weakly increasing and bounded, and
For each line, the series whose language be here together a convergent sum,

the limit of the sums of columns is equal to the sum of the series of time k is the limit of the box k (which is also his supremum) . The series has a convergent sum if and only if the (weakly increasing) sequence of line sums is bounded and therefore convergent.

As an example, consider the infinite series of line

<math> :: (1 +1 / n) ^ n = \ sum_ {k = 0} ^ {n} \ binom nk / n ^ k = \ sum_ {k = 0} ^ {n} \ {k frac1! } \ Times \ frac nn \ times \ frac {n-1} n \ times \ cdots \ times \ frac {nk +1} n </ math>

where n approaches maximal (the limit of this series is e). Here is the matrix entry in line n and column is k

<math> \ Binom nk / n ^ k = \ frac1 {k!} \ Time \ frac nn \ times \ frac {n-1} n \ times \ cdots \ times \ frac { nk +1} n; </ math>

the line ( security k) that is weakly increasing with n and bounded (by 1 /' k), when the line only finitely many nonzero terms, so pain is enough 2 for;'s Theorem now you include limits of line sums <math> (1 +1 / n) ^ n </ math> by taking the sum of the other restrictions, which <math> \ frac1 {k!} </ math>.

Lebesgue's monotone convergence Theorem

This generalizes Theorem past one, and is probably the most important monotone convergence Theorem. It also called Beppo Levi 's Theorem.

Theorem === === Let (' X, Σ, μ') will have a measurement location. Let <math> f_1, f_2, \ ldots </ math> be a pointwise non-Reduced sequence of [0, ∞]-valued Σ- Measurable functions, eg ' all' k ≥ 1 and all x in X,

Next, the pointwise limit of sequence <math> (f_ {n}) </ math> be f. That is, for every x in X,

<math> F (x): = \ lim_ {k \ to \ infty} f_k (x). \, </ Mathematics>

Then f is Σ- Measurable and

<math> \ Lim_ {k \ to \ infty} \ int f_k \, \ mathrm {d} \ mu = \ int f \, \ mathrm {d} \ mu. </ Mathematics>

'Remark. If the sequence <math> (f_k) </ math> satisfies the estimate' μ-almost everywhere, one can find some N ∈ Σ with μ ( N) = 0 such that the sequence <math> (f_k (x)) </ math> is non-Reduce all <math> x \ notin N </ math>. The advantage is that because all k,

<math> \ Int f_k \, \ mathrm {d} \ mu = \ int_ {X \ backslash f_k N} \, \ mathrm {d} \ mu, \ \ text {and} \ \ int f \, \ mathrm {d} \ mu = \ int_ {X \ backslash N} f \, \ mathrm {d} \ mu </ math>

provided that f is Σ-Measurable (see example <ref name="SCHECHTER1997"> Erik Schechter (1997). Analysis and Its origins. </ ref> section 21.38).

Proof === === We first show that the f is Σ- Measurable (see Example ^[1] section 21.3). Do this, it is enough to show that the inverse image of an interval [0, t'] in f is the element of Sigma algebra Σ on X, because the (closed) intervals to the Borel Sigma algebra for reals. I' = [0, t'] be such a subinterval of [0, ∞]. Storage

<math> F ^ {-1} (I) = \ {x \ in X \,). real test. theory of measure and integration. {{cite book}}: Check date values in: |year= (help); Text "\, f (x) \ in I \}. </ Mathematics>

Since I' is a prolonged and close <math> \ forall k, f_k (x) \ Le f (x) </ math>,

<math> F (x) \ in I \ Leftrightarrow f_k (x) \ in I, \ forall k \ in \ mathbb {N}. </ Mathematics>

Therefore,

<math> \ {X \ in X \," ignored (help); line feed character in |year= at position 26 (help)CS1 maint: extra punctuation (link) CS1 maint: year (link) \ {x \ in X \, | \, f_k (x) \ in I \}. </ Math>

Note that each set in the countable intersection is a element of Σ because it is the inverse image of a Borel subset in a Σ- Measurable Function <math> f_k </ math>. Since Sigma algebras is, by definition, closed in countable intersection, this indicates that f is Σ-Measurable. In general, the supremum of the countable family of functions is also Measurable Measurable.

Now we will prove the rest of the monotone convergence Theorem. The fact that f is Σ-Measurable implies that information <math> \ int f \, \ mathrm {d} \ mu </ math> is well defined.

We begin by showing that <math> \ int f \, \ mathrm {d} \ mu \ geq \ lim_k \ int f_k \, \ mathrm {d} \ mu. </ Mathematics>

By means of the Lebesgue integral,

<math> \ Int f \, \ mathrm {d} \ mu = \ sup \ {\ int g \, \ mathrm {d} \ mu \, | \, g \ in SF, \ g \ leq f \} </ math>

where SF is co-Measurable Σ simple functions on X. Since <math> f_k (x) \ leq f (x) </ math> of every x ∈ X, we have that

<math> \ Left \ {\ int g \, \ mathrm {d} \ mu \, | \, g \ in SF, \ g \ leq f_k \ right \} \ subseteq \ left \ {\ int g \, \ mathrm {d} \ mu \, | \, g \ in SF, \ g \ leq f \ right \}. </ Math>

Hence, since the supremum of a subset can not be serious than all set, we have that:

<math> \ Int f \, \ mathrm {d} \ mu \ geq \ lim_k \ int f_k \, \ mathrm {d} \ mu </ math>

and restrictions on political rights, since the sequence is monotonic.

We now prove the inequality in the other direction (which also follows from Fatou's Lemma), if we find that

<math> \ Int f \, \ mathrm {d} \ mu \ leq \ lim_k \ int f_k \, \ mathrm {d} \ mu. </ Mathematics>

It follows from the definition of the integral, with a non-Decreasing sequence (g _{''' k ) of non-not easy eight functions such that g k ≤ f and such that}

<math> \ Lim_k \ int g_k \, \ mathrm {d} \ mu = \ int f \, \ mathrm {d} \ mu. </ Mathematics>

It suffices to prove that for all <math> k \ in \ mathbb {N} </ math>,

<math> \ Int g_k \, \ mathrm {d} \ mu \ leq \ lim_j \ int f_j \, \ mathrm {d} \ mu </ math>

because if this is true for all k, then the limit of the left-hand side will be less than or equal to the right-hand side.

We will show that if g''' _{k is a simple and Function}

for every x, then

<math> \ Lim_j \ int f_j \, \ mathrm {d} \ mu \ geq \ int g_k \, \ mathrm {d} \ mu. </ Mathematics>

Since the integral is linear, we may violate the Function <math> g_k </ math> in the constant value, reduce to the case in which <Math > g_k </ math> is the illustration of a Function element B of Sigma algebra Σ. In this case, we assume that <math> f_j </ math> is a sequence of functions which Measurable supremum of each point of B is more than or equal to a a.

To prove this result, fix ε> 0 and denote the sequence of Measurable hours

<math> B_n = \ {x \ in B: f_n (x) \ geq 1 - \ epsilon \}. \, </ Mathematics>

By monotonicity of the integral, it says the following areas <math> n \ in \ mathbb {N} </ math>,

<math> \ Mu (B_n) (1 - \ epsilon) = \ int (1 - \ epsilon) 1_ {B_n} \, \ mathrm {d} \ mu \ leq \ int f_n \, \ mathrm {d} \ mu </ math>

From that <math> \ lim_j f_j (x) \ geq g_k (x) </ math>, the x in B in <math> B_n </ math> enough for high value of n, and therefore

<math> \ Bigcup_n B_n = B. </ Math>

Therefore, we have that

<math> \ Int g_k \, \ mathrm {d} \ mu = \ int 1_B \, \ mathrm {d} \ mu = \ mu (B) = \ mu \ left (\ bigcup_n B_n \ right) . </ Math>

Using the monotonicity property of the measure, we can continue the above equalities as follows:

<math> \ Mu \ left (\ bigcup_n B_n \ right) = \ lim_n \ mu (B_n) \ leq \ lim_n (1 - \ epsilon) ^ {-1} \ int f_n \, \ mathrm {d } \ mu. </ Mathematics>

Take' k → ∞, and the fact that this is true for any ε', the result follows.