$f(x)>0$ or $f(x)+f(x+1)>0$. Is it true $\int_{-\infty}^{\infty}f>0$ Suppose $f$ is a continuous function on $\Bbb R$ s.t $f(x)>0$ or $f(x)+f(x+1)>0$. Is it true $\int_{-\infty}^{\infty}f>0$?
My guess: It is true.
I was thinking in this way that we can partition $\Bbb R$ s.t $f^+=\{x \in \Bbb R: f(x)>0\}$ then $f^-=\{x \in \Bbb R: f(x)<0\}$ and $f^0=\{x \in \Bbb R: f(x)=0\}$.
Now observe for any point $x \in f^-$ we have $f(x+1)>-f(x)>0$. Now $\int_{-\infty}^{\infty}f=\lim_{n \to \infty}\int_{-n}^nf=\lim_{n \to \infty}\lim_{k \to \infty}R(P_{kn},f)$ where $P_{kn}$ is the partition of the interval $[-n,n]$ into intervals of length $\frac 1k$ and $R(P_{kn},f)$ is the Riemann Sum(observe one thing that $f$ being continuous is Riemann Integrable on $[-n,n]$. Moreover, we do not care whether $\lim_{n \to \infty}\int_{-n}^nf$ diverges or not. Even if it diverges we have to show that it takes $+\infty$. 
Now if $x\in f^- \cap [-n,n]$ then it will be in one of the subinterval $I_{kn} \subseteq [-n,n]$(WLOG we consider that $I_{kn} \subseteq f^-$ and corresponding $I_{kn}+1$ would be positive and the sum of these two will be positive.
I am writing my intuition. Please tell me whether the argument is correct or not or if there is a much more precise way to prove or write this problem!! 
 A: As far as I understand the question and from the clarification by MathematicsStudent1122, the statement is false.
Assumptions: $f$ is continuous and for every $x\in {\mathbb R}$, either $f(x)>0$ or $f(x)+f(x+1)>0$. 
The following counter example establishes that even the improper Riemann integral may fail to exist.
Counter example: Let $\epsilon>0$. Define $f(x)$ by $$f(x)=\left\{\begin{array}{ccc}\frac{\epsilon}{1+x^2},&&x\leq 0\\ -2^k\epsilon,&&x=\frac{6k-5}4,k=1,2,\cdots\\ \epsilon,&&\frac{3k-2}2\leq x\leq \frac{3k-1}2,k=1,2,\cdots\\ (2^k+1)\epsilon,&&x=\frac {6k-1}4\end{array}\right.$$ and extend by joining line segments to make the function continuous. Pictorially the graph for $x\geq 0$ consists of wavelike pattern of shape $V$, a horizontal segment and a $\Lambda$ of "period" $\frac 32$. By construction the integral $$\int_{-\infty}^0f(x)~dx =\left.{\epsilon}\arctan(x)\right|_{-\infty}^0=\frac{\pi}2\epsilon,$$ which is under control (i.e. convergent). For each of the $V$ shapes, the integral is $$\int_{\frac{3k-3}2}^{\frac{3k-2}2}f(x)~dx =-\frac 12\cdot\frac 12(2^k+1)\epsilon+\frac 12\epsilon=\frac 12\epsilon(\frac 12-2^{k-1}),k=1,2,\cdots.$$ The integral of the horizontal segments is $$\int_{\frac{3k-2}2}^{\frac{3k-1}2}f(x)~dx =\frac 12\epsilon,$$ and the integral of each of the $\Lambda$ shapes is $$\int_{\frac{3k-1}2}^{\frac{3k}2}f(x)~dx =\frac 12\cdot\frac 12\cdot 2^k\epsilon+\frac 12\epsilon=\frac 12\epsilon(1+2^{k-1}),k=1,2,\cdots.$$ It follows that the integral of one wave (consisting of $V$, $-$, and $\Lambda$) is $\frac 54\epsilon$, so the integral of the first $k$ waves is $$\int_0^{\frac {3k}2}f(x)~dx =\frac 54k\epsilon\rightarrow \infty.$$ But the integral of the first $k$ waves plus a $(k+1)$st $V$ shape is $$\int_0^{\frac {3k+1}2}f(x)~dx =\frac 54k\epsilon+\frac 12\epsilon(\frac 12-2^k)\rightarrow -\infty.$$ This shows that the improper Riemann integral does not exist.
However if one assumes that the Lebesgue integral (allowing $\pm \infty$) exists, then the assertion $$\int_{-\infty}^{\infty}f(x)~dx >0$$ is true. This can be argued as follows. Let $$E=\{x|f(x)\leq 0\}$$ and $$E_1=\{x+1|x\in E\}.$$ Then by assumptions, one has $$f(x)>0,\forall x\in E_1$$$$\Rightarrow E_1\subset E^c.$$ Let now $E_2=E^c\setminus E_1.$ Then $$\int f(x)~dx =\int_E f(x)~dx +\int_{E^c}f(x)~dx $$ $$=\int_E f(x)~dx +\int_{E_1}f(x)~dx +\int_{E_2}f(x)~dx $$ $$=\int_E f(x)~dx+\int_E f(x+1)~dx +\int_{E_2}f(x)~dx $$ $$=\int_E(f(x)+f(x+1))~dx +\int_{E_2}f(x)~dx >0.$$ QED
A: Define $B := \{x| f(x) \leq 0 \}$, $C := \{x| f(x-1) \leq 0 \}$; note that $C = B + 1$ and $B$ and $C$ are disjoint, closed and of equal measure. Note
$\int_{B \cup C} f = \int_B f + \int_{C} f $. Now we note that for all $d > 0$ if $B_d = B \cap[-d,d]$ and $C_{d} = C \cap [-d+1,d+1]$ 
$\int_{B_d}f = \lim_{n \rightarrow \infty}\sum_{k = 0}^{k = n-1}\int_{B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}]} f(b_{d,k,n})$
and
$\int_{C_d}f = \lim_{n \rightarrow \infty}\sum_{k = 0}^{k = n-1}\int_{C_d \cap [-d+1 + \frac{2dk}{n}, -d+1 + \frac{2d(k+1)}{n}]} f(c_{d,k,n})$
where $b_{d,k,n} = inf\{b|b \in B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}] \}$
and $c_{d,k,n} = inf\{c|c \in C_d \cap [-d+1 + \frac{2dk}{n}, -d+1 + \frac{2d(k+1)}{n}]\}$.
Now observe that $c_{d,k,n} = b_{d,k,n} +1$ and
$m(B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}]) = m(C_d \cap [-d+1 + \frac{2dk}{n}, -d+1 + \frac{2d(k+1)}{n}])$ thus we can write
$\int_{C_d} f = \lim_{n \rightarrow \infty}\sum_{k = 0}^{k = n-1}\int_{B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}]} f(b_{d,k,n} +1)$
Hence
$\int_{B_d}f + \int_{C_d}f = \lim_{n \rightarrow \infty}\sum_{k = 0}^{k = n-1}\int_{B_d \cap [-d + \frac{2dk}{n}, -d + \frac{2d(k+1)}{n}]} f(b_{d,k,n} +1)+f(b_{d,k,n}) > 0$
Hence 
$\int_{B}f + \int_{C}f > 0$ 
now $f > 0 $ on $(B \cup C)^c$ hence 
$\int f = \int_{B\cup C}f + \int_{(B\cup C)^c}f > 0 $. Note here that the integral of $f$ is assumed to exist.
