an real analysis problem of functions of improper integrals is the  statement true/false.
Let $f$ be a continuous integrable function of $\mathbb{R}$ such that either $f(x) > 0$ or
$f(x) + f(x + 1) > 0$ for all x$ \in \mathbb{R}$. Then $\int_{-\infty}^{\infty} f(x)dx>0$
is the above statement is true please help someone.
i am sorry that i could not wright it properly.the ingration will be from -infinity to infinity.thanks for help.
 A: The first case is obvious.
For the second, note that
\begin{eqnarray*}
\int_{-\infty}^{\infty} f(x)dx &=& \frac{1}{2} \int_{-\infty}^{\infty}f(x)+f(x)dx\\
&=& \frac{1}{2}\left[\int_{-\infty}^{\infty}f(x)dx + \int_{-\infty}^{\infty}f(x)dx\right]\\
&=& \frac{1}{2}\left[\int_{-\infty}^{\infty}f(x)dx + \int_{-\infty}^{\infty}f(y+1)dy\right]\\
&=& \frac {1}{2}\left[\int_{-\infty}^{\infty}f(x)+f(x+1)dx\right] > 0,
\end{eqnarray*}
where the third equality follows from the change of variables $y := x-1$ and the last inequality is a direct consequence of $f(x)+f(x+1) > 0 \ \ \forall x \in \mathbb{R}.$ 
A: If for example you have a particular function like this that is positive when $x>0$ and negative when $x<0$, just split the integral in two at zero.
Where $f(x)>0$, it's obvious.
For $\int_{-\infty}^{0} f(x)dx$ just integrate your inequality:
$$\int_{-\infty}^{0} f(x)dx+\int_{-\infty}^{0} f(x+1)dx>0$$
Then change variables u=x+1 and see that you now have $$2\int_{-\infty}^{0} f(x)dx>0$$.
the general case just need you to split $R$ as many times as $f$ changes its sign.
A: The first is obviously true. The second I feel uneasy with; to give a totally random example: $$f(x)=\sin\pi x+\frac{1}{2}$$
This fills the condition $f(x)+f(x+1)>0$ but the integral $\displaystyle\int_{-\infty}^{\infty} f(x)\;dx$ fails to exist 
(not sure if the word "integrable" disregards divergent improper integrals - please correct me if this is off-track)
