I would like to know if the following idea (for which i didn't find references) is correct or not.
Let $\mu_n$ be an exponentially tight sequence of non negative measure i.e. such that there exists a sequence of compact $K_t$ such that for any $n$ $$ \mathbb{P}(\mu_n \notin K_t) \le e^{-nt}$$ Suppose that we are in the case that for every test function $\phi\in C_c(\mathbb{R})$, $0<\mu_n(\phi)\le ||\phi||_{\infty}\mu_n(1)$, so I can easily study $\mu_n(1)$ which does not need to be finite.
My idea: the exponential tightness implies the uniform integrability
$$\mathbb{E}(\mu^n(1)\mid\mu^n(1)>k)=\int_k^\infty dt\mathbb{P}(\mu_n(1)>t)\le \int_k^\infty dt e^{-nt}=\frac{1}{n}e^{-nk}.$$
Therefore for every $\epsilon>0, \,\exists\, k\in[0,\infty)$ such that $$\sup_n\mathbb{E}(\mu^n(1)\mid\mu^n(1)>k)\le \sup_n \frac{1}{n}e^{-nk}<\epsilon$$
which gives the uniform integrability.
Is it correct or did I miss something?
I think your calculation of the conditional expectation is wrong. When you condition on the event that the random variable is larger than k, should the conditional expectation be larger than k as well? Your formula calculates the expectation of $\max[X-k, 0]$. Please see this post Find expected value using CDF.
I suppose what you need to calculate is $E[X1_{\{X\ge k\}}]$.