I'd like to know if there's a proof of Dunford-Pettis Theorem without using relatively advanced theorems of functional analysis such as Eberlein–Smulian Theorem. Since I'm only interested in probability spaces, I'd like a proof that only uses standard measure theory theorems which you can find, for instance, in "Measure and Integration" (by Salamon) and "Advanced Probability" (by Sokol and Rønn-Nielsen).

Below is the version of the Dunford-Pettis Theorem that I'm interest in.

Dunford-Pettis Theorem: Let $(\Omega,\Sigma,\mathbb{P})$ be a probability space and $\{X_n\}_{n\in\mathbb{N}}\subseteq L^1(\mathbb{P})$ be a collection of integrable random variables. Then the following propositions are equivalent:

1. Every sequence of $\{X_n\}_{n\in\mathbb{N}}$ has a further subsequence $\{X_{n_k}\}_{k\in\mathbb{N}}$ such that exists $X\in L^1(\mathbb{P})$ with $\mathbb{E}[X_{n_k}Y]\to \mathbb{E}[XY]$ as $k\to \infty$ for all bounded random variables $Y$.
2. $\sup_{n\in\mathbb{N}}\mathbb{E}[|X_n|]<\infty$ and for all $\varepsilon >0$ there's $\delta >0$ such that for all $E\in \Sigma $ with $\mathbb{P}(E)<\delta $ we have $\sup_{n\in\mathbb{N}}\mathbb{E}[\mathbf{1}_{E}|X_n|]<\varepsilon $

If you know a reference that contains a proof of the previous theorem which avoids relatively advanced theorems of functional analysis, please tell me. Also, if you know how to prove it, then please give me some hints because I don't know where I should start to prove that theorem.

Thank you for your attention!

I'd like to know if there's a proof of the Dunford-Pettis theorem without using relatively advanced theorems of functional analysis such as Eberlein–Smulian Theorem. Since I'm only interested in probability spaces, I'd like a proof that only uses standard measure theory theorems which you can find, for instance, in "Measure and Integration" (by Salamon) and "Advanced Probability" (by Sokol and Rønn-Nielsen).

Below is the version of the Dunford-Pettis theorem that I'm interest in.

Dunford-Pettis theorem: Let $(\Omega,\Sigma,\mathbb{P})$ be a probability space and $\{X_n\}_{n\in\mathbb{N}}\subseteq L^1(\mathbb{P})$ be a collection of integrable random variables. Then the following propositions are equivalent:

1. Every sequence of $\{X_n\}_{n\in\mathbb{N}}$ has a further subsequence $\{X_{n_k}\}_{k\in\mathbb{N}}$ such that exists $X\in L^1(\mathbb{P})$ with $\mathbb{E}[X_{n_k}Y]\to \mathbb{E}[XY]$ as $k\to \infty$ for all bounded random variables $Y$.
2. $\sup_{n\in\mathbb{N}}\mathbb{E}[|X_n|]<\infty$ and for all $\varepsilon >0$ there's $\delta >0$ such that for all $E\in \Sigma $ with $\mathbb{P}(E)<\delta $ we have $\sup_{n\in\mathbb{N}}\mathbb{E}[\mathbf{1}_{E}|X_n|]<\varepsilon $

If you know a reference that contains a proof of the previous theorem which avoids relatively advanced theorems of functional analysis, please tell me. Also, if you know how to prove it, then please give me some hints because I don't know where I should start to prove that theorem.

Thank you for your attention!

I'd like to know if there's a proof of Dunford-Pettis Theorem without using relatively advanced theorems of functional analysis such as Eberlein–Smulian Theorem. Since I'm only interested in probability spaces, I'd like a proof that only uses standard measure theory theorems which you can find, for instance, in "Measure and Integration" (by Salamon) and "Advanced Probability" (by Sokol and Rønn-Nielsen).

Below is the version of the Dunford-Pettis Theorem that I'm interest in.

Dunford-Pettis Theorem: Let $(\Omega,\Sigma,\mathbb{P})$ be a probability space and $\{X_n\}_{n\in\mathbb{N}}\subseteq L^1(\mathbb{P})$ be a collection of integrable random variables. Then the following propositions are equivalent:

1. Every sequence of $\{X_n\}_{n\in\mathbb{N}}$ has a further subsequence $\{X_{n_k}\}_{k\in\mathbb{N}}$ such that exists $X\in L^1(\mathbb{P})$ with $\mathbb{E}[X_{n_k}Y]\to \mathbb{E}[XY]$ as $k\to \infty$ for all bounded random variables $Y$.
2. $\sup_{n\in\mathbb{N}}\mathbb{E}[|X_n|]<\infty$ and for all $\varepsilon >0$ there's $\delta >0$ such that for all $E\in \Sigma $ with $\mathbb{P}(E)<\delta $ we have $\sup_{n\in\mathbb{N}}\mathbb{E}[\mathbf{1}_{E}|X_n|]<\varepsilon $

If you know a reference that contains a proof of the previous theorem which avoids relatively advanced theorems of functional analysis, please tell me. Also, if you know how to prove it, then please give me some hints because I don't know where I should start to prove that theorem.

I also have another question: In the item 1 of the previous theorem, can I replace "for all bounded random variables $Y$" by "for all continuous bounded random variables $Y$" without breaking that equivalence?

Thank you for your attention!

I'd like to know if there's a proof of Dunford-Pettis Theorem without using relatively advanced theorems of functional analysis such as Eberlein–Smulian Theorem. Since I'm only interested in probability spaces, I'd like a proof that only uses standard measure theory theorems which you can find, for instance, in "Measure and Integration" (by Salamon) and "Advanced Probability" (by Sokol and Rønn-Nielsen).

Below is the version of the Dunford-Pettis Theorem that I'm interest in.

Dunford-Pettis Theorem: Let $(\Omega,\Sigma,\mathbb{P})$ be a probability space and $\{X_n\}_{n\in\mathbb{N}}\subseteq L^1(\mathbb{P})$ be a collection of integrable random variables. Then the following propositions are equivalent:

1. Every sequence of $\{X_n\}_{n\in\mathbb{N}}$ has a further subsequence $\{X_{n_k}\}_{k\in\mathbb{N}}$ such that exists $X\in L^1(\mathbb{P})$ with $\mathbb{E}[X_{n_k}Y]\to \mathbb{E}[XY]$ as $k\to \infty$ for all bounded random variables $Y$.
2. $\sup_{n\in\mathbb{N}}\mathbb{E}[|X_n|]<\infty$ and for all $\varepsilon >0$ there's $\delta >0$ such that for all $E\in \Sigma $ with $\mathbb{P}(E)<\delta $ we have $\sup_{n\in\mathbb{N}}\mathbb{E}[\mathbf{1}_{E}|X_n|]<\varepsilon $

If you know a reference that contains a proof of the previous theorem which avoids relatively advanced theorems of functional analysis, please tell me. Also, if you know how to prove it, then please give me some hints because I don't know where I should start to prove that theorem.

Thank you for your attention!