Is there a integrable function non-vanishing a.e.

Question

Is there a lebesgue integrable function that is non-vanishing a.e.? In general, given any $\sigma$-finite measure space, is there a way to define a integrable function that is non-vanishing a.e?

Answer 1

Take $f(x)=\exp(-x^2)$. Then $\|f\|_{L^1(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)}=\sqrt{\pi}<+\infty$ and $f(x)>0$ for every $x\in\mathbb{R}$.

Answer 2

Well I could take just a really cheap counterexample to the general case. Let $X = \{0\}$, and put a measure on $X$ by setting $\mu \{0\} = \infty$. So you might ask whether it is true for $\sigma$-finite spaces (I suspect it is, but I'll have to think about how to prove it).

EDIT: Okay, how about this? Suppose $X$ is $\sigma$-finite with measure $\mu$, so it is a countable union $A_1 \cup A_2 \cup \cdots$ of subsets of (nonzero) finite measure. We can arrange the sets so that WLOG, $A_1 \subseteq A_2 \subseteq A_3 \subseteq \cdots$.

Now let $\delta_k = \mu A_k$. Since $X$ is the disjoint union of $A_1, A_2 \setminus A_1, A_3 \setminus A_2$ etc. we can define a measurable function $f$ on $X$ by letting $f$ have the value $\frac{1}{2^{k-1} \delta_k}$ on $A_k \setminus A_{k-1}$ for $k = 1, 2, ...$ (let $A_0 = \emptyset$).

Then $$\int\limits_X f d\mu = \sum\limits_{k=1}^{\infty} \int\limits_{A_k \setminus A_{k-1} } f d\mu = \sum\limits_{k=1}^{\infty} \frac{1}{2^{k-1} \delta_k} \mu(A_k \setminus A_{k-1}) \leq \sum\limits_{k=1}^{\infty} \frac{1}{2^{k-1}\delta_k} \mu(A_k)$$ $$ = \sum\limits_{k=1}^{\infty} \frac{1}{2^{k-1}} < \infty$$