Uniform $L^1$ convergence of a double sequence of random variables.

Let $X_n^m$ be a double sequence of integrable random variables such that $X_n^m \rightarrow_n X^m$ in probability, where $X^m$ is another sequence of integrable random variables. Assume that such convergence is uniform in $m$ and that $X_{n}^m \rightarrow_n X^m$ in $L^1$.

Does it follow that $X_{n}^m \rightarrow_n X_m$ in $L^1$ uniformly in $m$?

Thanks

It does not follow from the assumptions: take $X_{m,n}:=mn\cdot \chi_{(0,n^{-2})}$ on $(0,1)$ endowed with Lebesgue measure. We have for $\varepsilon\lt 1$ and each integer $n$, $$\sup_{m\in\mathbb N}\mathbb P\{|X_{m,n}|\gt \varepsilon \}=n^{-2},$$ hence there is uniform convergence in probability to $0$. We also have $\mathbb E[X_{m,n}]=\frac mn$, hence we have convergence in $\mathbb L^1$ to $0$, but this convergence is not uniform in $m$.