The answer to the question asked is no, i.e. the Haar integral $I(f)$ of a nonzero, nonnegative continuous function $f$ is always positive. See Hewitt & Ross, Theorem (15.5)(i).15.5)(i).

The mistake in your argument is that $f(x)=1/(1+x^2)$ is not almost periodic. Indeed, recall that $f$ is almost periodic iff $\forall\varepsilon>0$ $\exists L>0$ such that every interval of length $L$ contains an $\varepsilon$-almost period of $f$, i.e. a number $P$ such that $\sup_{\mathbf{R}}|f(\cdot - P)-f(\cdot)|\leqslant\varepsilon$. But it is clear (from looking at the graph) that $\sup_{\mathbf{R}}|f(\cdot - P)-f(\cdot)|>\frac12$ for every $P$ in every interval $[10, 10+L]$ say.

Re: "just think of a continuous, bounded, AP integrable function", there is no such function $f$ except zero. Indeed if $\int_{-\infty}^\infty |f(x)|dx<\infty$ then $I(|f|)=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T |f(x)|dx=0$, whence $f=0$ by the above-quoted Theorem (15.5).

2 fixed integration limits

The answer to the question asked is no, i.e. the Haar integral $I(f)$ of a nonzero, nonnegative continuous function $f$ is always positive. See Hewitt & Ross, Theorem (15.5)(i).

The mistake in your argument is that $f(x)=1/(1+x^2)$ is not almost periodic. Indeed, recall that $f$ is almost periodic iff $\forall\varepsilon>0$ $\exists L>0$ such that every interval of length $L$ contains an $\varepsilon$-almost period of $f$, i.e. a number $P$ such that $\sup_{\mathbf{R}}|f(\cdot - P)-f(\cdot)|\leqslant\varepsilon$. But it is clear (from looking at the graph) that $\sup_{\mathbf{R}}|f(\cdot - P)-f(\cdot)|>\frac12$ for every $P$ in every interval $[10, 10+L]$ say.

Re: "just think of a continuous, bounded, AP integrable function", there is no such function $f$ except zero. Indeed if $\int_{-\infty}^\infty |f(x)|dx<\infty$ then $I(|f|)=\lim_{T\to\infty}\frac1{2T}\int_{-\infty}^\infty I(|f|)=\lim_{T\to\infty}\frac1{2T}\int_{-T}^T |f(x)|dx=0$, whence $f=0$ by the above-quoted Theorem (15.5).

1

The answer to the question asked is no, i.e. the Haar integral $I(f)$ of a nonzero, nonnegative continuous function $f$ is always positive. See Hewitt & Ross, Theorem (15.5)(i).

The mistake in your argument is that $f(x)=1/(1+x^2)$ is not almost periodic. Indeed, recall that $f$ is almost periodic iff $\forall\varepsilon>0$ $\exists L>0$ such that every interval of length $L$ contains an $\varepsilon$-almost period of $f$, i.e. a number $P$ such that $\sup_{\mathbf{R}}|f(\cdot - P)-f(\cdot)|\leqslant\varepsilon$. But it is clear that $\sup_{\mathbf{R}}|f(\cdot - P)-f(\cdot)|>\frac12$ for every $P$ in every interval $[10, 10+L]$ say.

Re: "just think of a continuous, bounded, AP integrable function", there is no such function $f$ except zero. Indeed if $\int_{-\infty}^\infty |f(x)|dx<\infty$ then $I(|f|)=\lim_{T\to\infty}\frac1{2T}\int_{-\infty}^\infty |f(x)|dx=0$, whence $f=0$ by the above-quoted Theorem (15.5).