Is there a function $f:[0,\infty) \to \{-1, 1\}$ such that $\int_0^\infty{f(x)\,dx}$ is well defined?

Question

I suspect it is impossible to find such an $f$, but I can see one way it might be possible to as well.

If it is possible to construct two sets $A$ and $B$ which partition the non-negative reals such that $A \cap [0,a)$ and $B \cap [0,a)$ both have measure $a/2$ for any positive real $a$, then take $f^{-1}(1) = A$, $f^{-1}(-1) = B$. We would have $\int_0^a{f(x)\,dx}=0$ for all $a$ and $\lim_{a \to \infty}\int_0^a{f(x)\,dx}=0$.

Answer 1

It is possible as an improper integral. Fix a conditionally convergent series $\sum_{n=1}^{\infty} a_n$, and let $T_n = \sum_{k=1}^{n} |a_k|$. Then define $f$ as

$$ f(x) = \operatorname{sgn}(a_n) \qquad\text{whenever}\quad x \in [T_{n-1}, T_n). $$

Then it follows that

$$ \int_{T_{k-1}}^{T_k} f(t) \, \mathrm{d}t = \int_{T_{k-1}}^{T_k} \operatorname{sgn}(a_k) \, \mathrm{d}t = \operatorname{sgn}(a_k)(T_k - T_{k-1}) = a_k. $$

So, whenever $x \in [T_n, T_{n+1})$, we get

\begin{align*} \left| \int_{0}^{x} f(t) \, \mathrm{d}t - \sum_{k=1}^{n} a_k \right| = \left| \int_{T_n}^{x} f(t) \, \mathrm{d}t \right| \leq a_{n+1}, \end{align*}

and this converges to $0$ as $n \to \infty$. Therefore,

$$ \int_{0}^{\infty} f(t) \, \mathrm{d}t = \sum_{k=1}^{\infty} a_k. $$

Answer 2

Choose a sequence $u_n \ge 0$ such that $u_n \to 0$ but $\sum u_n = \infty$ nonetheless, for example $u_n = 1/n$. Then, starting at $0 \in [0,\infty)$, define $f$ as $+1$ on an interval of length $u_1$, then as $-1$ on an interval of length $u_1$ again; then repeat, $+1$ for a length of $u_2$ and $-1$ for a length of $u_2$, etc. Such an $f$ solves your problem for the improper Riemann/Lebesgue integral, as deviations of the integral from $0$ will be under $u_n \to 0$. Formally, with the partial sums $S_n = \sum_{k<n} u_k$: $$ f(x) = \begin{cases} +1 & \text{for } 2S_n \le x < 2S_n + u_n , \\ -1 & \text{for } 2S_n + u_n \le x < 2S_{n+1} . \end{cases} $$

Answer 3

For $x\geqslant0$, define $$f(x) := \begin{cases} 1 & \text{if }\; 2^x-\lfloor 2^x\rfloor<\frac12, \\ -1 & \text{otherwise}. \end{cases}$$ Then $\int_0^xf(t)\,\mathrm dt$ tends to a limit exponentially fast as $x\to\infty$.

Edit note: The initial version was unnecessarily complicated.