Given a bounded sequence in $L^1([a,b])$, is $(t\mapsto \int_a^t f_n \,\mathrm d\lambda)_n$ equicontinuous? We are given a bounded sequence $(f_n)_{n\in \mathbb N}$ in $L^1([a,b])$.
This means there is some $M>0$ such that for all $n\in\mathbb N$, $\int_{[a,b]} |f_n|\,\mathrm d\lambda \leq M$.
I wonder whether the corresponding sequence $(F_n)_{n\in\mathbb N}$ of integrals
$$F_n : [a,b]\to \mathbb R,\quad t\mapsto \int_a^t f_n \,\mathrm d\lambda$$
is equicontinuous at all $x\in [a,b]$,
i.e. for each $\varepsilon > 0$ there exists some $\delta> 0$ such that for all $n\in\mathbb N$ and $y\in [a,b]$ with $|y-x|<\delta$ we have $|F_n(x)-F_n(y)| < \varepsilon$.
Edit: Clarifcation. I don't assume the functions $f_n$ to be bounded in the sense that there is some $M$ with $f_n(t)\leq M$ for all $t\in [a,b]$.
Instead, I assume that all $f_n$ are bounded in the metric space $(L^1([a,b]), \lVert \cdot \rVert_{L^1})$ as explained above.
 A: No, consider $[a,b]=[0,1]$ and $f_n(x) = \begin{cases} n - n^2x & \text{if} & x \le \frac{1}{n} \\ 0 & \text{otherwise} \end{cases}$
Then obviously $$\|f_n\|_{L_1} = \int_0^{\frac{1}{n}} n -n^2x dx = 1 - n^2 \int_0^{\frac{1}{n}}xdx = 1-\frac{1}{2} = \frac{1}{2} $$ so $\{f_n\}_n$ is bounded in $L_1([0,1])$. However, looking at $x=0$, and taking $y \in (0,1)$ we have $$ |F_n(0) - F_n(y)| = \int_0^y n - n^2x dx = ny - \frac{n^2y^2}{2}$$
At $y = y(n) = \frac{1}{n}$ you get $|F_n(0)-F_n(y)| = \frac{1}{2}$. So it is true that for $\varepsilon = \frac{1}{2}$ there is no $\delta > 0$ (because sequence $y(n) = \frac{1}{n}$ converges to $0$) such that for all $n$ and all $y < \delta$ you have $|F_n(0)-F_n(y)| < \varepsilon$.
A: Since $|F_n(x) - F_n(y)| = | \int_x^y (f_n)| \leq \int_x^y | f_n| \leq (b-a) M,$ yes.
A: YES.
If $g_n(x)=\int_a^x f_n(t)\,dt$ and $\sup_{x\in[a,b]}|f(x)|=M$, then
$$
|g_n(x)-g_n(y)|=\left|\int_{[x,y]}f(t)\,dt\right|\le 
\int_{[x,y]}|f(t)|dt\le M|x-y|.
$$
Hence, for every $\varepsilon>0$,
$$
|x-y|<\delta=\varepsilon/M\quad\Longrightarrow\quad
|g_n(x)-g_n(y)|<\varepsilon.
$$
