Let $\mathcal M$ be a tight family of probability measures on a separable metric space. Let $f\in \bigcap_{\mu \in \mathcal M}L^1(\mu)$. For each $\mu\in \mathcal M$, define $\mathcal M_0 := \{ f\mu\}_{\mu \in \mathcal M}$.
Is $\mathcal M_0$ tight? It certainly is if $f$ is bounded. Is it true in general? If not, what assumptions on $\mathcal M$ or $f$ would make $\mathcal M_0$ tight?
Let $\{X_n\}$ be a sequence of integrable random variable converging almost surely (to $X$) but not in $L^{1}$. Let $\mathcal M =\{P\circ X_n^{-1}: n \geq 1\}$. This family is tight because $P\circ X_n^{-1} \to P\circ X^{-1}$ weakly. Let $f(x)=|x|$. Then tightness of $\{fdP_n:n \geq 1\}$ is equivalent to uniform integrability of $\{X_n\}$ which is not true (because $\{X_n\}$ does not converge in mean).