finite non-negative integral implies integral of the square is finite

Question

Given $f:(\mathbb{R},\mathbb{B}(\mathbb{R})) \rightarrow (\mathbb{R}, \mathbb{B}(\mathbb{R}))$ is a non-negative function, and given $f$ is bounded, i'm trying to show that $\int_{\mathbb{R}}f d\mu < \infty \Rightarrow \int_{\mathbb{R}}f^{2}d \mu < \infty$

Notation: $\mathbb{B}(\mathbb{R})$ denotes the Borel-sigma algebra on the real line. $d \mu$ is the Lebesgue measure.

What I've tried:

I might be going down the wrong road completely, but my idea was, to say that since $\int_{\mathbb{R}}fd\mu <\infty$ then we know that $\int_{\mathbb{R}}f^{+}d\mu - \int_{\mathbb{R}}f^{-}d\mu$ are both finite. Where $f^{+}(t):=\max(f(t),0)$ and $f^{-}(t):=\max(-f(t),0)$.

If we want $\int_{\mathbb{R}}f^{2}d\mu < \infty$ then by definition we require $\int_{\mathbb{R}}(f^{+})^{2}d\mu$ and $\int_{\mathbb{R}}(f^{-})^{2}d\mu$ to be finite. If I evaluate these and get a finite answer then I'm finished?

Answer 1

It is given that $f \geq 0$. $\int f^{2}d\mu \leq M\int fd\mu<\infty$ (where $M$ is such that $f(x) \leq M$ for all $x$).

Answer 2

$\int f<\infty$ can only be true if $f$ is integrable, i.e. if $f$ is measurable and $\int |f|<\infty$.

If $|f|\leq M$ then $f^2\leq M|f|$ so that: $$\int f^2\leq\int M|f|=M\int|f|<\infty$$