Krdytkyu

Let $f$ be a continuous integrable function on $mathbb{R}$ such that for each $x$, either $f(x)>0$ or $f(x)+f(x+1)>0$. Then is $int_{-infty}^{infty}f(x)dx>0$ ?

Intuitively, I think this is true, because whenever on a set $A$ , $f(x)<0$ , then $int_{Acup A+1} f(x)dx>0$. But I do not know how to write this argument rigorously.

Here is a sketch, note that the integrals below make sense since $A,B$ are open and $C$ is closed.

Let $A:={ x : f(x) < 0 }$ and $B= { x +1 :x in A }$. Then, both $A,B$ are open and disjoint*( since $f(x)+f(x+1)>0$).

Let $C:= mathbb R backslash (A cup B)$. Then $A,B,C$ are pairwise disjoint, $A,B$ open $C$ closed (i.e. they are Borel if you are familiar with Borel sets) and hence
$$
int_{mathbb R} f(x) d x = int_A f(x) d x+ int_B f(x) d x +int_C f(x) d x \
stackrel{t=x-1}{===}int_A f(x) d x+ int_A f(t+1) d t +int_C f(x) d x \
=int_A left( f(x) +f(x+1) right)d x +int_C f(x) d x \
$$

Now, each integral is $geq 0$ thus $int_{mathbb R} f(x) dx geq 0$.

If $A neq emptyset$ since $A$ is open and $f(x)+f(x+1)>0$ it is easy to argue using continuity that $int_A f(x)+f(x+1) dx >0$.

If $A =emptyset$ then $C =mathbb R$ and $f(x) geq 0$ for all $x$. Now, it is easy to argue that $f$ cannot be identically zero, and by the same argument $int_C f(x) dx >0$.

Let $A:={xinBbb R: f(x)>0}$. Because $f$ is continuous then $A$ is open and piecewise-connected (by "piecewise-connected" I mean that it is the union of countable disjoint open intervals).

WLOG suppose that $AneqBbb R$. This imply that $A^complement$ is the union of countable and closed intervals. Then $A^complement=bigcup_{jin I} R_j$, for $|I|lealeph_0$, where $R_jcap R_k=emptyset$ for $jneq k$ and each $R_j$ is a closed interval.

Now note that because $f(x)le 0$ for each $xin A^complement$ then we knows that for each $R_j$ there is some (non-empty) open interval $I_j$ such that $lambda(I_j)gelambda(R_j)$, $sup R_j=inf I_j$ and $f|_{I_j}>0$. From here its easy to see that

$$Bbb R=I_0cupleft(bigcup_{jin I}(R_jcup I_j)right)$$

is a disjoint partition of $Bbb R$ where $I_0$ could be empty, that is, each interval $I_j$ or $R_j$ is disjoint of any other interval $I_k$ or $R_k$. Hence

$$int_{Bbb R} f(x),lambda(dx)=\=int_{I_0} f(x),lambda(dx)+sum_{jin I}int_{R_j}(f(x)+f(x+1)),lambda (dx)+sum_{jin I}int_{H_j}f(x),lambda(dx)$$

where $H_j:=I_jsetminus(R_j+1)$.

Then clearly $int_{R_j} (f(x)+f(x+1)),lambda(dx)ge 0$ for each $jin I$ because $f(x)+f(x+1)>0$ and $lambda(R_j)ge 0$. Also $int_{H_j} f(x)lambda(dx)ge 0$ because $lambda(H_j)ge 0$ and $H_jsubset A$. By last, if $I_0neqemptyset$ then $lambda(I_0)>0$ and $I_0subset A$.

Then adding all we find that the integral is necessarily positive, because if some $lambda(R_j)=0$ (that is, if $R_j$ is a singleton) then $lambda(H_j)>0$, and viceversa.