Show that $\int _{-\infty}^\infty f(x) dx>0 $ . Let $f$ be a continuous and integrable funtion on $\Bbb R$ such that either $f(x)>0$ or $f(x)+f(x+1)>0$ for all $x$.

Show that $$\int _{-\infty}^\infty f(x) dx>0 .$$

If I assume that $f(x)>0$ then I will have $$\int _{-\infty}^\infty f(x) dx>0 .$$
But I am unable to proceed for $f(x)+f(x+1)>0$ for all $x$.
Please give some hints
 A: Hint:
$$
\int_{-\infty}^\infty f(x+1) dx=\int_{-\infty}^\infty f(x) dx 
$$
A: Note that if $f(x) < 0$ then $f(x+1) > |f(x)| 0$.  This allows us to define the function
$$
g(x) = \left\{ \begin{array}{ccl} f(x) &\mbox{if}& f(x-1) > 0 \wedge f(x) > 0 \wedge f(x+1) > 0\\ 
\frac{f(x+1)+f(x)}{2} &\mbox{if}& f(x)<0 \mbox{ implying } f(x-1) > 0 \wedge f(x+1) > -f(x) \\
\frac{f(x-1)+f(x)}{2} &\mbox{if}& f(x-1)<0 \mbox{ implying } f(x) > -f(x-1) > 0 \end{array}
\right.
$$
These three conditional cover all cases.  
Now compare $\int g(x)$ to $\int f(x)$.  For each range $(a,b)$ of $x$ in which $g(x) - f(x) = p(x) > 0$ (because $f(x) < 0$), there is a corresponding region $(a+1,b+1)$ in which $f(x) > g(x) = p(x)$.  (This statement depends on continuity.)  So 
$$
\int_{-\infty}^\infty g(x) = \int_{-\infty}^\infty f(x) 
$$ 
But $g(x)$ is everywhere positive, so 
$$
 \int_{-\infty}^\infty f(x) = \int_{-\infty}^\infty g(x) > 0 
$$ 
A: For x between 0 and 1, it is obvious that Sum {f(x+/-n)}>0. But the integral is just the average over all x between 0 and 1 so it is also greater than 0.
