Why is the zero extension of an $L^p$ function in $L^p$?

Question

Let $u \in L^p(0,1)$. Define $\tilde u:(0,\infty) \to \mathbb{R}$ as the function which equals $u$ on $(0,1)$ and $\tilde u =0$ on $(1,\infty)$.

I cannot figure out why this function is measurable. The integral is obviously finite but I don't know about the measurability. How do I justify it?

Answer 1

Look at the inverse image of a generating measurable set.

$$\bar{u}^{-1}(a,b)$$

If $a<0<b$ then because inverse images and unions commute, i.e. $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ we get:

$$\bar{u}^{-1}(a,b)=\bar{u}^{-1}(a,0)\cup \bar{u}^{-1}(\{0\})\cup\bar{u}^{-1}(0,b)$$

$$=u^{-1}(a,0)\cup (1,\infty)\cup u^{-1}(\{0\})\cup u^{-1}(0,b)$$

this is a finite union of measurable sets, hence measurable.

If $0\not\in (a,b)$ then $\bar{u}^{-1}(a,b)=u^{-1}(a,b)$ which, again, is measurable. So in both cases the inverse image of a generating set for the (Borel) $\sigma$-algebra is measurable, hence the extended function is.

Answer 2

The function $f$ is measurable if and only if there exists a sequence of step functions that converge to $f$ almost everywhere.

Your $u$ is integrable, hence measurable. Therefore you have that sequence of step functions with support in $(0,1)$ which converges to $u$ a.e. Note that the same sequence converges a.e. to $\tilde u$ on $(0,\infty)$, hence $\tilde u$ is measurable.