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Let $\mathbb{R}_{\geq0}$ denote the set of non-negative reals and $+\infty$, and $\mathbb{Z}^+$ denote the set of positive integers. I will also let $\lambda$ denote the Lebesgue measure on $\mathbb{R}$ .

Let there be a function $T:\mathbb{R}_{\geq0}^{\mathbb{Z}^+}\rightarrow \mathbb{R}_{\geq0}$ such that:

  1. For any functions $c:\mathbb{Z}^+\rightarrow \mathbb{R}_{\geq0},x:\mathbb{Z}^+\times\mathbb{Z}^+\rightarrow \mathbb{R}_{\geq0}$, we have:$$T((\sum_{k=1}^{\infty}c_kx(n,k))_{n\in \mathbb{Z}^+ })\leq\sum_{k=1}^{\infty}c_kT((x(n,k))_{n\in \mathbb{Z}^+})$$
  2. For any sequence of non-negative Lebesgue measurable functions $\{f_n\}_{n\in \mathbb{Z}^+}$, the function $T(\{f_n\}_{n\in \mathbb{Z}^+})$ that sends $x$ to $T((f_n(x))_{n\in \mathbb{Z}^+})$ is Lebesgue measurable.

An example of a function $T$ that satisfies these conditions is the function that sends $(x_n)_{n\in Z^+}$ to $\sum_{n\in Z^+}x_n$.

Question: Does it follow that $$T((\int_{\mathbb{R}}f_n\,d\lambda)_{n\in \mathbb{Z}^+})\leq \int_{\mathbb{R}}T(\{f_n\}_{n\in \mathbb{Z}^+})\,d\lambda$$ I am intrested to know the answer to this question.

I am also very interested to see any counterexamples (if there are any).

Thank you.

Amr
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    I am a bit messed up with the notation, but does it follow directly that the condition on the infinite sums implies that the integral inequality holds for simple functions, all having the same partition? – SBF Jan 08 '13 at 17:53
  • The inequality for infinite sums along with the measurability of $T({f_n}_{n\in Z^+}$ implies the inequality for integral. – Amr Jan 08 '13 at 20:48
  • that's something you want to check, isn't it? Or was it a respond to my comment? In the latter case, unfortunately, I cannot see how does it respond – SBF Jan 09 '13 at 07:17
  • It was a respond. After reading your first comment again, I realized that I don't get it. (What partition, btw ?) – Amr Jan 09 '13 at 08:07
  • Say, if for all $n$ you have $$ f_n(x) = \sum\limits_{k\in \Bbb Z^+}f_{nk}\cdot 1_{E_k}(x) $$ where $(E_k)_{k\in \Bbb Z^+}$ is some fixed measurable partition of $\Bbb R$, does the integral inequality hold? – SBF Jan 09 '13 at 08:52
  • Perhaps some examples (of such functions $T$) would clarify the question a little bit. – Ewan Delanoy Jan 09 '13 at 09:19
  • @Ewan Delanoy An example of a function $T$ is the function that sends $(x_n){n\in Z^+}$ to $\sum{n\in Z^+}x_n$ – Amr Jan 09 '13 at 09:51
  • I am not sure if there exists such functions $T$ which are not linear (or more precisely, restrictions of linear functions). Do you have examples in mind? – Siméon Jan 09 '13 at 11:37
  • @Ju'x Another example of the function $T$ is the function that sends $(x_n){n\in Z^+}$ to $\lim,\sup{n\rightarrow\infty}x_n$ – Amr Jan 09 '13 at 12:45
  • @Ilya Yes.${}{}{}{}{}{}$ – Amr Jan 09 '13 at 13:44
  • @Amr: since $f_n$ are measurable functions, then there is a sequence $f_{n,k_n}$ of simple functions converging to each of them. As a result, by taking a finite number of limits w.r.t. $k_n$ in the inequality which holds for simple functions, we obtain that it holds for their limits. Thus, your assertion shall be true at least for the case when all but a finite number of ${f_n}$ are simple. – SBF Jan 09 '13 at 14:11
  • @Ilya I don't see how this works. Could you post it as an answer and explain all details ? – Amr Jan 09 '13 at 20:45
  • @Amr: Well, I can right the proof for the simple functions over the same partition. However, for taking the limit some form of commutativity between $T$ and the monotonic limit may be needed, and that may not be in the general case. – SBF Jan 10 '13 at 09:08
  • @Ilya Yes Yes it might be easy to do it for simple functions having the same partition, however this is still very far from answering the question. But I 'd like to see your thoughts so you might post whatever you have in mind as an answer. – Amr Jan 10 '13 at 09:20
  • An exemple. Suppose $T$ is convex and $\big(\sum_{k=1}^{\infty}\cdot c_k\cdot x(k,n)\big)_{n\in\mathbb{Z}^+}$ a convex combination of the set ${ x(k,n) }$. – Elias Costa Jan 10 '13 at 11:05
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    @Elias yes there are other examples as well. $(x_n){n\in Z^+}\rightarrow \sup{n\in Z^+}x_n$ – Amr Jan 10 '13 at 11:12

1 Answers1

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As requested, there are some thoughts on the topic. If we have to pass from the properties of the sum to that of the Lebesgue integral, it's always worth to show that the property we are interested in does hold at least for the simple functions.

What does hold. Assume that $\{E_j\}_j$ is a countable measurable partition of $\Bbb R$ and consider a collection of non-negative functions $\{f_i\}_i$ $$ f_{i,j}(x) = \sum_{j} a_{ij}\cdot1_{E_j}(x) $$ so that all functions are simple and are given on the same partition. Note that $f:=T(\{f_i\}_i)$ is $$ f(x) = \sum_{j} T(\{a_{ij}\}_i)\cdot 1_{E_j}(x) $$ as you can check when substitute each single $x\in E_j$ into $T(\{f_i\}_i)$. As a result we have $$ \int f_i\mathrm d\lambda = \sum_j a_{ij} \lambda(E_j), \quad \int f\mathrm d\lambda = \sum_j T(\{a_{ij}\}_i) \lambda(E_j) $$ and if we put $x_{ij} = a_{ij}$ and $c_j = \lambda(E_j)$ then by the property of $T$ that is given to us: $$ T\left(\left\{\int f_i\mathrm d\lambda\right\}_i\right) = T\left(\left\{\sum_j a_{ij} \lambda(E_j)\right\}_i\right)\leq\sum_j \lambda(E_j) T(\{a_{ij}\}_i) = \int T(\{f_i\}_i)\mathrm d\lambda. $$

What holds in addition. If we are initially given simple functions on different (finite) partitions, then starting from $(\{E^i_j\}_j)_i$ - a countable collection indexed by $i\in \Bbb Z^+$ of finite partitions $E^i = \{E_{ij}\}_j$, there exists some countable partition $\{F_k\}_k$ which agrees with all of them. To show this, we first note that there is a countable partition $F^{1,2}$ underlying both $E^1$ and $E^2$, thus it holds for any finite collection $J\subset \Bbb Z^+$ of countable partitions - let us call the corresponding underlying partition by $F^J$. Now, if I am correct, $F$ contains at most as many elements as $\bigcup_n F^{1,\dots,n}$ and the latter is countable. Thus, our previous assumption is not restrictive and hence your claim does hold under assumption you have provided for all collections of simple functions.

What does hold under additional assumptions. Now, to pass from simple functions to measurable ones we can use the pointwise approximation by simple functions as in the construction of the Lebesgue integral. However, that requires interchanging $\sup_n$ and $T$ in the following way:

if $\{(y^n_i)_n\}_i$ is a countable collection of monotonically non-decreasing sequences $y_i = (y_i^0,y_i^1,\dots)$, then $$ \lim_n T(\{y_i^n\}_i) = T(\{\lim_n y^n_i\}_i). \tag{1} $$ Under such condition your inequality shall also hold for arbitrary measurable functions. I am not sure, whether $(1)$ is already implied by the assumptions raised in OP, so if you want to find a counterexample, I guess you shall look into cases when $(1)$ does not hold.

By the way, under the assumptions on the additivity of $T$ a condition similar to $(1)$ is completely characterized, and is known to hold exactly for transition kernels on $\Bbb R$, see [Proposition 1.3, Chapter 1] in Revuz, "Markov Chains".

That's pretty much what I could say, so hopefully it helps a bit.

A positive kernel on $\Bbb R$ is a function $K:\Bbb R\times \mathfrak B(\Bbb R)\to [0,\infty)$ such that $K(x,\cdot)$ is a positive measure for all $x$ and such that $K(\cdot,B)$ is a measurable function for all measurable $B$. Let $\mathfrak B_+(\Bbb R)$ denote the space of all non-negative measurable functions. Then $K$ defines an operator on that space by $$ Kf(x) = \int f(y)K(x,\mathrm dy). $$

Proposition: (Revuz) An additive and homogeneous operator $V$ on $\mathfrak B_+(\Bbb R)$ is associated with some positive kernel if and only if for every increasing sequence $\{f_n\}\subset \mathfrak B_+(\Bbb R)$ it holds that $$ V(\lim_n f_n) = \lim_n V(f_n). $$

SBF
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  • Thank you for sharing your thoughts, it would be great if you can include proposition 1.3 in your answer. – Amr Jan 10 '13 at 11:57
  • I think that there are functions T that satisfy the conditions of my question but do not satisfy statement (1) of your answer – Amr Jan 10 '13 at 11:59
  • @Amr: agree, that's why I pointed out that even additive functions $T$ may not satisfy $(1)$ if they are not kernels – SBF Jan 10 '13 at 12:04
  • I see . I missed that one. – Amr Jan 10 '13 at 12:05
  • @Amr: Added, though as I wrote it does not apply directly - just looks at similar properties. I guess, you can find a counterexample - try to play around with $T$ defined via $\sup_i$. – SBF Jan 10 '13 at 12:11