# Is space complete and separable?

$$\displaystyle M =\{x\in\mathbb{R}^{(0,1]} : x \text{ is continuous on } (0,1] \text{ and } x(t) = o(\ln{}t)\ (t \rightarrow +0)\}$$. We will consider metric $$\displaystyle\rho(x, y) = \int\limits_{0}^{1}|x(t)-y(t)|dt$$.

1. Is $$(M,\rho)$$ complete space?
2. Is $$(M,\rho)$$ separable space?
3. If $$(M,\rho)$$ is not complete find its completion.

I've considered $$\displaystyle x_k(t) = t^{\frac{1}{k}}\ln{}t$$, $$x_k \in M$$. Obviously, $$x_k \rightarrow \ln{}t\notin M$$, as $$\rho(x_k,\ln{}t) = 1- \displaystyle\frac{1}{(\frac{1}{k} + 1)^2} \rightarrow 0.$$ In the same way we can show that it is Cauchy sequence. So, $$(M, \rho)$$ is not complete.

But I've stuck with two other questions. I think that completion of $$M$$ is $$\mathbb{L}_1(0,1]$$. How can one show it? If it is so, we will also have an answer for the second question. $$\mathbb{L}_1(0,1]$$ is separable, so $$M$$ is separable.

Integrability of $$\ln t$$ shows that $$M$$ is a subset of $$L^{1}(0,1]$$.
If $$x \in C(0,1]$$ then $$x_i(t)=x(t)$$ for $$t \geq \frac 1 i, x_i(t)=itx_(\frac 1i)$$ for $$0 defines a sequence in $$M$$ converging to $$x$$ in $$L^{1}(0,1]$$. Combined with the fact that continuous functions are dense in $$L^{1}(0,1]$$ it follows that $$M$$ is dense in $$L^{1}(0,1]$$. This implies that $$L^{1}(0,1]$$ is the completion of $$M$$.