I have a following question. For $t\in[0,T]$, let $f_t:\mathbb{R}\rightarrow\mathbb{R}$ be uniformly integrable family i.e. $\{f_t,\,t\in[0,T]\}$- uniformly integrable. We consider the space of such families $\mathcal{U}$. We put a norm $\lVert f_t\rVert=\sup_{t\in[0,T]}\int_{\mathbb{R}}|f_t(x)|\,dx$. Is the space $(U,||\cdot||)$ a complete space?
My idea: Let us take $f^n_t$- a Cauchy sequence from $\mathcal{U}$ i.e. $\lVert f^n_t-f^m_t\rVert\rightarrow 0$.
We have that, for any $t\in[0,T]$, $\int_{\mathbb{R}}|f^n_t(x)-f^m_t(x)|\,dx\le \lVert f^n_t-f^m_t\rVert\rightarrow 0$.
Hence, for all $t\in[0,T]$, $f^n_t$ is a Cauchy sequence in $L^1$. Hence, since it is a complete space, there exists $f_t$ such that $\int^T_0|f^n_t(x)-f_t(x)|\,dx\rightarrow 0$. But, the question is if $\lVert f^n_t-f_t\rVert\rightarrow 0$?
For all $\varepsilon>0$ there exists $t\in[0,T]$ such that $\lVert f^n_t-f_t\rVert\le \int_{\mathbb{R}}|f^n_t(x)-f_t(x)|\,dx+\varepsilon$. This completes the proof.
Is there any mistake?
Uniform integrability of $(f_t)$ is equivalent to finiteness of $\sup_t \int |f_t(x)|dx$ together with the following: for every $\epsilon >0$ there exists $\delta >0$ such that $\int_A |f_t(x)| dx <\epsilon$ whenever $m(A) <\delta$.
Choose $n$ such that $\int |f_t^{n}(x)-f_t(x)|dx <\epsilon$ for all $t$. By uniform integrability of $(f^{n}_t)$ we get $\delta$ such that $\int_A |f^{n}_t(x)| dx <\epsilon$ whenever $m(A) <\delta$. Hence $\int_A |f_t(x)| dx <2\epsilon$ whenever $m(A) <\delta$. This proves uniform integrability of $(f_t)$.