If $f \in L^{1}(0,\infty)$ and $x \in (0, \infty)$, then $xf \in L^{1}(0,\infty)$? Suppose $f \in L^{1}(0,\infty)$. Is it true that
$$
\int_{0}^{\infty}xf(x)dx < \infty \ \ 
$$
 A: Note: This answer concerns the original formulation of the question. The other answer (by JonathanZ) concerns the revised question.

Given $\sigma$-finite measure space $(\varOmega, \mathcal A, \mu)$ and $p \in [1, \infty)$, the space $L^p(\varOmega, \mathcal A, \mu)$ is a vector space, which implies that yes, $L^1(0, \infty)$ is closed under scalar multiplication.
(It's actually a quotient vector space, which I mention in case you are new to the notion of $L^p$ spaces. This detail is not germane to your question concerning closure under scalar multiplication.)
A: No. $f(x) = \min (1, 1/x^2)$ is a counterexample.
As $f$ is positive everywhere, we can just check
$$ \lim_{a \to + \infty} \int_0^a f(x)dx.$$
So long as $a \gt 1$
$$\begin{align}
\int_0^a f(x)dx &= \int_0^1 1dx + \int_1^a x^{-2}dx \\
&= 1 + (-a^{-1} - (-(1^{-1}))\\
&=2 - \frac{1}{a}
\end{align}$$
so $f\in L^1(0,\infty)$.
But
$$\begin{align}
\int_0^a xf(x)dx &= \int_0^1 xdx + \int_1^a x^{-1}dx \\
&= 1/2 + (\ln(a) - \ln(1))\\
&=\ln(a) - 1/2
\end{align}$$
so $xf\notin L^1(0,\infty)$.
