Linear Space and Inner Products Prove that the set $$ V=\{f\quad|\quad f:\mathbb{R} \to \mathbb{R}, f \quad\text{is absolutely integrable over} \quad \mathbb{R} \} $$ is a linear space over $\mathbb{R}$.
Is it necessary to go over and show (explicitly) that all conditions for linear space are satisfied, or can I simply argue that since $V$ is the set of absolutely integrable function, the integrals have finite values, and the rest follows from the linearity of integrals?

Let $V=\{f:\mathbb R\rightarrow \mathbb R~|~~\int_\mathbb R |f(x)|dx <\infty \}$. 
You need to check that


*

*$\forall \lambda\in \mathbb R$ and $f\in V$, then $\lambda f\in V$

*$\forall f,g\in V$ then $f+g\in V$.


Let $f,g$ and $\lambda$ be as above. The first condition is easy to prove: the function $\lambda f$ is, by definition, given by
$(\lambda f)(x):=\lambda f(x)$; so
$$\int_\mathbb R |(\lambda f)(x)|dx=\int_\mathbb R |\lambda f(x)|dx= |\lambda|\int_\mathbb R |f(x)|dx<\infty.$$
The second condition is proven by using the triangular inequality for the absolute value $|\cdot|$ and linearity of the integral:
$$\int_\mathbb R |( f+g)(x)|dx=\int_\mathbb R |f(x)+g(x)|dx\leq \int_\mathbb R (|f(x)|+|g(x)|)dx= \int_\mathbb R |f(x)|dx +\int_\mathbb R|g(x)|dx <\infty.$$
