Lebesgue integral - convergence of supremums

Question

Let $\newcommand{\d}{\mathrm{d}}(f_{a, n})_{a \in (0, 1), \, n \in \mathbb{N}} \subseteq L^1$ and $f_{a} \in L^1$ for every $a\in (0,1)$. Suppose that for each parameter $a$ one has that

$$ f_{a, n} \stackrel{n\to\infty}{\longrightarrow} f_a.$$

Assume further that there exist constants $C_1, C_2$ such that for all $a \in (0, 1), \, n \in \mathbb{N}$

$$ C_1 \leq \int f_{a, n} \,\d\mu\leq C_2. $$

Is it then true that

$$ \sup_{a \in (0, 1)} \int f_{a, n}\,\d\mu \stackrel{n\to\infty}{\longrightarrow} \sup_{a \in (0, 1)} \int f_a \,\d\mu ? $$

Answer 1

Let us consider $L^1(0,1)$ with the Lebesgue measure $\newcommand{\d}{\mathrm{d}}\lambda$. Then we can define

\begin{align} f_{a,n} := \begin{cases} 0, & \text{if}\; a \notin \{2^{-i}: i \in \mathbb{N}\}, \\ 1, & \text{if there exists an}\; i \in \mathbb{N} \;\text{sucht that}\; a=2^{-i} \;\text{and}\; n\leq i. \end{cases} \end{align}

For each $a\in (0,1)$ we have that $\{f_{a,n}\}_{n\in\mathbb{N}}$ converges in $L^1$-norm to the zero-function. This means that

$$\sup_{a\in (0,1)}\int f_a \,\d \lambda=0.$$

On the other hand if we fix $n\in \mathbb{N}$, we obtain:

$$\sup_{a\in (0,1)}\int f_{a,n}\,\d\lambda=1.$$

Answer 2

Thanks!

In this setting, could we at least infer that

$$ \sup_{a \in (0, 1)} \int f_a \, d\mu \leq \sup_{a \in (0, 1)} \int f_{a, \, n} \, d\mu \, ?$$