A question involving weak convergence in $\sigma(L^1, L^\infty)$

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Exercise 4.15 from H. Brezis - "Functional analysis..."

Let I = (0, 1) and $f_n = n e^{-n x}$ a sequence of functions. Show that $$ f_n \;\; \text{does not converge weakly to} \;\; 0 \;\; \text{in} \;\; \sigma (L^1, L^\infty). $$

In the book, at Solutions, we have the following:

Note that $\int f_n \varphi \to 0, \;\; \forall \varphi \in C_c(I)$. Suppose, by contradiction, that $f_{n_k} \rightharpoonup f$ weakly $\sigma(L^1, L^\infty)$. It follows that $$\int f \varphi = 0 \;\; \forall \varphi \in C_c(I).$$

Why $\int f \varphi = 0 \;\; \forall \varphi \in C_c(I)$?

Thank you!