I'm learning about measure theory and $L^P$ spaces and need help with the following questions:
True or False (justify):
$(1)$ Let $f:(-1, 1) \to \mathbb{R}$ measurable on $(-n, n), \; \forall n \in (0, 1)$. Then $f$ is measurable on $(-1, 1)$.
$(2)$ If $f, g \in L^2 \implies fg \in L^2$.
$(3)$ If $f \in L^2 \big( [1, M] \big), \; \forall M > 1 \implies f \in L^2 \big( [1, +\infty) \big)$.
My work:
$(2)$ The proposition is false. Let
$$f(x) = \frac{1}{x^{\frac{1}{4}}}\chi_{[0, 1]} = g(x).$$
Then
$$fg(x) = \frac{1}{x^{\frac{1}{2}}} \chi_{[0, 1]} = \frac{1}{\sqrt x} \chi_{[0, 1]}.$$
Therefore
$$\int_\mathbb{R} \left|fg(x)\right|^2\, dx = \int_\mathbb{R} \left(\frac{1}{\sqrt x}\right)^2 \chi_{[0, 1]} \, dx = \int_{[0, 1]} \frac{1}{x} \, dx = +\infty.$$
so $fg \notin L^2$ (since the integral is divergent).
However
$$\int_\mathbb{R} \left|f\right|^2=\int_\mathbb{R} \left|g\right|^2 = \int_\mathbb{R} \left(\frac{1}{x^{\frac{1}{4}}}\right)^2 \chi_{[0, 1]} = \int_{[0, 1]}\frac{1}{\sqrt{x}} = 2$$
so $f, g \in L^2$ (since the integral is convergent).
Is my work correct for $(2)$? Are there any other counterexamples that are more straightforward then the above one?
Questions $(1)$ and $(3)$ are confusing me because they both appear extremely simple. My guess is that they are both true. For $(1)$, substituting $n=1$ we immediately get that $f$ is measurable on $(-1, 1)$. For $(3)$, since $M$ can reach infinity then it must be that $f \in L^2 \big( [1, +\infty) \big)$. I think I'm missing some key points for questions $(1)$ and $(3)$. Any help would be appreciated.
(1) is true, as $f$ is a limit of sequence of functions $f_n$, $f_n = f \cdot \chi_{(-1 + 1/n, 1-1/n)}$, which are all measurable on $(-1, 1)$ by assumption, so $f$ is measurable on $(-1, 1)$ as a limit of functions measurable on $(-1, 1)$. It's also not hard to show that it's measurable directly. For (3), see my comment above. Your argument for (2) looks correct.