Is this subspace of $L^1(\mathbb{R},m)$ closed?

Question

Let $K$ be the subspace of $L^1(\mathbb{R},m)$ which contains precisely the functions such that $\int f=0$. Is $K$ closed?

(EDIT: When I asked this question, I could only see that ${f:||f||_1=0}$ is closed.)

Answer 1

Let $f_n$ be a sequence of functions such that $f_n\rightarrow f$ in the $L^1$ norm. $L^1$ is complete so $f\in L_1$ Then

$$\left|\int fdm\right|\leq \left|\int (f-f_n)dm\right|+\left|\int f_ndm\right|\leq \|f-f_n\|_1\rightarrow 0$$

so $K$ is closed.

Answer 2

The function $f \mapsto \int_{\mathbb{R}} f$ is continuous on $L^1$ because

$$\left | \int_{\mathbb{R}} f - \int_{\mathbb{R}} g \right | \leq \int_{\mathbb{R}} |f-g| = \| f - g \|_{L^1}.$$

Now your set is the preimage of the closed set $\{ 0 \}$ under this function, so it is closed.