Let $u_n:\Omega\to\mathbb{R}$ be a given uniformly bounded sequence in $L^1(\Omega)$, where $\Omega$ is a bounded open subset $\mathbb{R}^N$.
Then there exists a bounded measure $\mu$ such that $$\int_{\Omega}u_n\,\psi\to\int_{\Omega}\psi\,d\mu$$ for all simple functions $\psi$.
I am not getting how to prove the above argument. Can anyone please help me how to prove it?
Thanks in advance.
It looks like the answer to the question (with the modification that the sequence may not converge but a sub-sequence does - see comments) follows immediately from a well known result in functional analysis.
Let's begin from the start, suppose we have a Banach space $X$ (in your case $X=L^1(\Omega)$). One can construct the dual space $X^\star$ of $X$ to be the set $$X^\star : = \{ \varphi:X\rightarrow\mathbb{C} : \varphi \text{ is a continuous functional}\}$$
In the case where $X=L^1(\Omega)$ Riesz theorem gives a general form for a function on $X$. By Riesz theorem every function $\varphi:L^1(\Omega)\rightarrow\mathbb{C}$ takes the form $\varphi(f) = \int fg d\mu$ for some $g\in L^\infty(\Omega)$. In other every functional of $L^1(\Omega)$ corresponds to an element in $L^\infty(\Omega)$. The other direction is also true, given a function in $g\in L^\infty(\Omega)$ the map $f\mapsto \int fg d\mu$ is a continuous functional of $L^1(\Omega)$.
A norm on $X^\star$: Elements in $X^\star$ are linear functions $\varphi:X\rightarrow \mathbb{C}$ and so you can consider the norm
$$\|\varphi\| = \sup_{\{x : \|x\|=1\}} \|\varphi(x)\|$$
The weak$^\star$ topology on $X^\star$ : The space $X^\star$ is clearly a vector space. Moreover, every element $x\in X$ correspond to a function of $X^\star$ (in some sense $X$ embedded in $(X^\star)^\star$) by $x (\varphi) = \varphi(x)$ (where $x\in X$ and $\varphi\in X^\star$). Let's call these functionals "simple functionals". The weak$^\star$ topology on $X^\star$ is defined to be the intersection of all typologies on $X^\star$ with respect to which all simple functionals are continuous.
We can finally state the theorem from which you can derive the answer to your question
Using this theorem your question becomes very easy: Indeed, we define functions $\mu_n:L^1(\Omega)\rightarrow S^1$ by $$\mu_n (\psi) = \int u_n\psi d\mu$$
Then $$\|\mu_n\| = sup_{\{\psi : \|\psi\|_{L^1}=1\}} \int \left|u_n \psi \right|d\mu$$
Since $\int |u_n \psi| d\mu \leq \|u_n\|_\infty \int |\psi| d\mu=\|u_n\|_\infty \|\psi\|_1 $ we conclude that $\|\mu_n\| = \|u_n\|_\infty$ which is totally bounded, say by $M$ but without loss of generality we can take $M=1$.
We conclude that $\mu_n \in B$, hence there exists a convergence sub-sequence in the weak$^\star$ -topology. In particular there exists $\mu\in B$ such that $\mu_{n_k}\rightarrow \mu$ with respect to that topology.
Again by the Riesz theorem this means that there exists $u\in L^\infty(\Omega)$ such that $\mu(\psi) = \int u \psi d\mu$.
Finally you remind yourself what is the weak$^\star$-topology on $X^\star$. It means that for all $x\in X$ the functions $x(\varphi) = \varphi(x)$ is continuous. In particular it means that for all $x\in X$ we have $x(\mu_{n_k})\rightarrow x(\mu)$ and so $\mu_{n_k}(x)\rightarrow \mu (x)$. Recall that $X=L^1(\Omega)$ and so we can take $x=\psi$ for some simple function $\psi$. We have that $\mu_{n_k}(\psi)\rightarrow \mu(\psi)$ which is the desired result.