Finding an increasing sequence of Lebesgue integrable functions such that supremum and integral cannot be interchanged

116 Views Asked by At

I am working in the setting of the monotone convergence theorem and looking for a case where the theorem cannot be applied.

So let $(X, A, \mu)$ be a measure space.

I want a sequence $(u_j)_{j \in \mathbb{N}} \subset L^1(\mu)$ an increasing sequence of integrable functions $u_1 \le u_2 \le ...$ with limit $u := \sup_{j \in \mathbb{N}}$. Such that $\sup_{j \in \mathbb{N}} \int u_j d \mu \not = \int \sup_{j \in \mathbb{N}} u_j d \mu$.

My def of $L^1$ is the set of all functions for which the positive and negative part integrate to a finite number.

Note: I only found examples reagrding the dominant convergence theorem, but not in the setting of the monotone convergence theorem.

1

There are 1 best solutions below

2
On BEST ANSWER

I am afraid that such a sequence does not exist.

Define $f_j=u_j+v$ where $v:=\max(-u_1,0)$ and $f=u+v$.

Then the monotone convergence theorem can be applied on these nonnegative functions, leading to:$$\lim_{k\to\infty}\int u_k\;d\mu+\int v\;d\mu=\lim_{k\to\infty}\int f_k\;d\mu=\int f\;d\mu=\int u\;d\mu+\int v\;d\mu\tag1$$

Here $v$ is the negative part of $u_1$ so that $\int v\;d\mu<\infty$.

Then $(1)$ can only be true if: $$\lim_{k\to\infty}\int u_k\;d\mu=\int u\;d\mu$$