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Let $A$ be a closable linear operator on $L$ with domain $D(A)$ and $\bar{A}$ be its closure. Then if $f_n \in D(A)$ is $\lim_n f_n$, provided it exists, in $D(\bar{A})$?

I encountered this question from a proof of a Proposition from Ethier and Kurz' Markov Processes. In the below proof, in the second line and the line right above (3.6), the authors state that $\int_{\epsilon}^{t_0} e^{-t}u(t)dt \in D(\bar{A})$ and $\int_0^\infty e^{-t}u(t)dt \in D(\bar{A})$. But from the hypotheses of the Proposition, we only assume that each $u(t)\in D(A)$. As the integrals are Riemann here, being the limit of vectors in $D(A)$, I thought they exist in $D(\bar{A})$ as $\bar{A}$ is a closed operator. However, I have not been able to prove this or find this statement anywhere else. What is the precise reasoning of this? I would greatly appreciate any help.

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    The closure $\bar{A}$ of $A$ is defined to be the minimal closed linear extension of $A$; more specifically, it is the closed linear operator $B$ whose graph is the closure (in $L \times L$) of the graph of $A$. – 2017-02-22
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    The [definition](https://en.wikipedia.org/wiki/Unbounded_operator#Closed_linear_operators) of $T$ is closed : if $f_n \in D(T), f_n\to f$ **and** $Tf_n \to g$ then $f \in D(T)$ and $g = T(f)$. Here you have $\overline{A}$ closed, $f_n \in D(\overline{A})$ and $f_n \to f$ but you need $\overline{A}f_n \to g$ for saying that $f \in D(\overline{A})$. – 2017-02-22

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