Linked Questions

5
votes
2answers
600 views

Dominated Convergence Theorem with “Almost Surely” replaced by “Convergence in Probability” [duplicate]

I want to show that if $\{X_n\}$ is a sequence of random variables such that: (1) $\exists X$ (measurable) such that $X_n \xrightarrow{P} X$ (2) $\exists Y$ with $E(|Y|) < \infty$ such that $|X_n|...
1
vote
0answers
227 views

Lebesgue's dominated convergence theorem also holds replacing “almost everywhere convergence” by “convergence in measure” [duplicate]

Possible Duplicate: Generalisation of Dominated Convergence Theorem Ive just read this on wikipedia: "$(X, M, μ)$ - measure space. If $\mu$ is $\sigma$-finite, Lebesgue's dominated convergence ...
4
votes
2answers
116 views

Convergence of functions [duplicate]

Assume that $(X,M,\mu)$ is a $\sigma$-finite space. Suppose that $|f_n|\leq g\in L^+$ and $f_n\rightarrow f$ in measure. Show that $\int f=\lim_{n\rightarrow\infty}\int f_n$. I tried taking a ...
1
vote
0answers
116 views

Convergence in measure instead of almost everywhere convergence in DCT [duplicate]

Let $(X, M ,\mu)$ be an arbitrary measure space. Dominated convergence theorem requires some sequence of integrable functions to converges to some function f almost everywhere. However, if $f^n$ ...
0
votes
0answers
84 views

Dominated convergence theorem in case of converge in measure. [duplicate]

I have heard that the dominated convergence theorem hold if almost everywhere convergence is replaced by convergence in measure. I concur if fn converges to f in measure then there exists a ...
14
votes
3answers
7k views

Sufficient condition for convergence of a real sequence [duplicate]

Let $(x_n)$ be a sequence of real numbers. Prove that if there exists $x$ such that every subsequence $(x_{n_k})$ of $(x_n)$ has a convergent (sub-)subsequence $(x_{n_{k_l}})$ to $x$, then the ...
10
votes
2answers
5k views

Measure Convergence Version of Lebesgue Dominated Convergence Theorem

I want to prove that LDCT(Lebesgue Dominated Convergence Theorem) continues to hold if I replace the hypothesis $f_n \to f$ (convergence pointwise) with $f_n\to f$ (convergence in measure): $$\int ...
7
votes
1answer
1k views

Dominated convergence and $\sigma$-finiteness

I am curious about the Dominated Convergence Theorem for a sequence of functions that converges in measure. Theorem: Let $(X,\mathcal{S},\mu)$ be a measure space. If $\{f_n\}, f$ are measurable, ...
4
votes
1answer
656 views

Variant of dominated convergence theorem

There are several variants of dominated convergence theorem. The standard one requires $f_n \to f$ a.e. and $|f_n|\leq g$ a.e. where $g$ is integrable. It can be weakened to only convergent in ...
4
votes
2answers
249 views

Question from Folland on modes of convergence

I have been reading through Folland, and I am having a hard time answering the following question. Any help will be much appreciated. Suppose $\lvert f_n \rvert \leq g \in L^1$ and $f_n \rightarrow ...
1
vote
2answers
308 views

Expected Value of Product of Sequences of Random Variables

I am trying to prove the following: Let $X_{n}$ be a sequence of random variables converging in probability to some random variable $X$. Furthermore $P(|Xn|>k)=0$ for all n and some $k>0$. ...
3
votes
1answer
194 views

Combining convergence in probability and the means of the positive sequence of r.v. implies convergence in L 1

Let $\{X_n\}$ be a collection of positive random variable with $X_n \rightarrow X$ in probability. Prove that if $E(X_n) \rightarrow E(X)$, then $X_n \rightarrow X$ in $L^1$. My partial answer: Let $(...
2
votes
2answers
100 views

$\lim_{n \to \infty} P(A_n) = 0$ implies $\lim_{n\to\infty}\int_{A_n}{X}dP = 0$

Let $(\Omega, \mathcal{A}, P)$ be a probability space and let $X$ be a non-negative, real random variable with $\int{X}dP < \infty$. If $A_n \in \mathcal{A}$ is a sequence such that $\lim_{n \to \...
2
votes
1answer
132 views

Show that $f_n1_{A_n}$ convergences in mean

Consider the measurable space $(\Omega,\mathcal{A},\mu)$. Let $f,f_1,f_2,\ldots$ be measurable functions on that measurable space and $A,A_1,A_2,\ldots\in\mathcal{A}$. Let $(f_n)$ converge in mean to $...
0
votes
1answer
51 views

If $X_n\overset{p}{\rightarrow}X$ and $\mathbb E(X_n)\rightarrow\mathbb E(X)$ then $\mathbb E|X_n−X|\rightarrow0$

For $X_n>0$ almost surely, show that if $X_n\overset{p}{\rightarrow}X$ and $\mathbb E(X_n)\rightarrow\mathbb E(X)$, then $\mathbb E|X_n−X|\rightarrow0$ My try: My idea was to consider the positive ...

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