Closed set in Banach?

a) Is true because $f\mapsto \int_{\mathbb{R}}f$ is a continuous linear functional on $L_1(\mathbb{R})$ . If we denote by $T$ this continuous mapping the space $L$ is
$T^{-1}(0)$ .

b) Is false. To prove it just observe that there are dense subspaces in $L_2(\mathbb{R})$ formed by integrable functions and with integral 0. Look at the Haar system Haar wavelet - Wikipedia, the free encyclopedia.

If the set of functions with integral 0 were closed, this would imply that the closure of the span of the Haar system, that is the whole $L_2(\mathbb{R})$ , would be included in $L$ , that certainly is not true.