Does $\int_0^{\infty} f(x) \, dx = \int_0^1 f(x) \, dx + \int_1^{\infty} f(x) dx$? Short version:
See title. I've checked a couple of analysis textbooks and Wikipedia quickly but haven't found an answer.
Long version:
I am trying to solve Exercise 4.1.3 from Probability Theory: A Comprehensive Course, by Klenke.

Let $1 \leq p' < p \leq \infty$ and let $\mu$ be $\sigma$-finite but not finite. Show that $\mathcal L^p(\mu) \setminus \mathcal L^{p'}(\mu) \neq \emptyset$.

I decided to work in $(1, \infty)$ with the Lebesgue measure $\lambda$, and consider the function $f(x) = \frac{1}{x}$. I already showed that $\int f \, d\lambda = \infty$ because the integrals of simple functions end up like the harmonic series (minus the first term).
I am trying to show that $\int f^2 \, d\lambda < \infty$. The simple function approach seems very unwieldy so I decided to apply the following result (abridged):

Let $f \colon \varOmega \to \mathbb R$ be measurable and $f \geq 0$ almost everywhere. Then $\int f \, d\mu = \int_0^{\infty} \mu\left(\{f \geq t\}\right) \, dt$.

I calculate that $\{ f \geq t \} = (1, 1/\sqrt t)$ for $0 \leq t \leq 1$, and $\emptyset$ for $t > 1$. This leads me to the integral
\begin{align*}
\int_0^{\infty} \lambda\bigl(\{x \in (1, \infty) \colon f(x) \geq t\}\bigr) \, dt = \int_0^1 \frac{1}{\sqrt t} - 1 \, dt + \int_1^{\infty} 0 \, dt = 1
\end{align*}
which seems to give me the right answer, but I'm not sure what to cite to justify breaking up an improper integral like that.
 A: If $f(x)$ and $g(x)$ are integrable functions on a measure space $(X,\mu)$, then one consequence of linearity of integration is that
$$
\int_X \big(f + g\big)\,d\mu = \int_X f\,d\mu + \int_X g\,d\mu.
$$
In particular, if $f(x)$ is an integrable function on $[0,\infty)$, then
$$
\int_{[0,\infty)}f(x)\,dx = \int_{[0,\infty)}f(x)(\mathbf1_{[0,1)}+\mathbf 1_{[1,\infty)})\,dx = \int_{[0,1)}f(x)\,dx + \int_{[1,\infty)}f(x)\,dx.
$$
Linearity is one of the fundamental properties of Lebesgue integration. Since you mentioned "improper integral" in your post, I will note that even in the Riemann theory of integration, if $\int_a^\infty f$ exists as an improper Riemann integral, and $a < b$, then it follows very straightforwardly from the definition of $\int_a^\infty f$ as the limit $\lim_{M\to\infty}\int_a^Mf$ that
$$
\int_a^\infty f = \int_a^b f + \int_b^\infty f.
$$
In particular, $\int_b^\infty f$ exists as an improper Riemann integral as well.
A: Even though my question arises from a measure-theoretic probability theory textbook, fundamentally it's a question about improper Riemann integration.
Put simply, for $c \in (a, b)$,
\begin{align*}
\int_a^{\infty} f(x) \, dx &= \lim_{b \to \infty} \int_a^b f(x) \, dx\\
&= \lim_{b \to \infty} \bigl(F(b) - F(a)\bigr)\\
&= \lim_{b \to \infty} \Bigl( \bigl(F(b) - F(c)\bigr) + \bigl(F(c) - F(a)\bigr) \Bigr)\\
&= \lim_{b \to \infty} \bigl(F(b) - F(c)\bigr) + \bigl(F(c) - F(a)\bigr)\\
&= \lim_{b \to \infty} \bigl(F(b) - F(c)\bigr) + \int_a^c f(x) \, dx\\
&= \int_c^{\infty} f(x) \, dx + \int_a^c f(x) \, dx
\end{align*}
so the improper integral can be broken up as desired.
