Let $g$ be a function from $(0, 1)$ to $\mathbb{R}$ and $\int_0^1 g^2(x)dx$ finite. Does this imply that $\int_0^1 g(x)dx$ is finite? Let $g$ be a function from $(0, 1)$ to $\mathbb{R}$ and $\int_0^1 g^2(x)dx$ finite. Does this imply that $\int_0^1 g(x)dx$ is finite? 
So I know that $| \int_0^1 g(x)dx | \leq \int_0^1 |g(x)|dx = \int_0^1 \sqrt{g^2(x)}dx$. Not sure how to continue. 

Let $A$ be a non-measurable subset of $[0,1]$ and put
$$g(x):=\cases{1\quad & $(x\notin A)$ \cr -1 & $(x\in A)$\cr}\ .$$
Then $g$ is not even integrable, but $\int_0^1 g^2(x)\ dx=1$. Therefore we better assume $g$ measurable to begin with. Under these circumstances
$$\left(\int_0^1 |g(x)|\cdot 1\ dx\right)^2\leq \int_0^1 g^2(x)\ dx \int_0^1 1^2\ dx<\infty$$
by Schwarz' inequality and assumption. This implies the existence of the integral $\int_0^1 g(x)\ dx$.
