non-negative continuous function where a sum of sample values diverges but integral converges Does there exist a non-negative continuous function $f:\mathbb{R}_+ \rightarrow \mathbb{R}_+$ which has a convergent integral:
$$ \int_0^\infty f \, dx < \infty \tag{1}$$
with the restriction that $\forall x \in \mathbb{R}_+, f(x) <1$. But, which has a strictly increasing sequence $ \{q_n\}_{n=0}^\infty \subset \mathbb{R}_+$ such that:
$$\sum_{n=0}^\infty f(q_n)=\infty \tag{2}$$
Note: My motivation for asking this question comes from probability, the Borel-Cantelli theorem in particular. 
 A: Start with the function $f(x)=0$  $\forall x\in\mathbb R$, then add a "triangle" to the graph with height $1$ and basis $1/2$ centered around $x=1$, repeat this process adding a triangle with height $1$ and basis $2^{-n}$, centered around $n$ for each positive integer $n$.
The integral will be $\sum\limits_{n=1}^\infty \frac{1}{2^n}$ which converges.
But $\sum\limits_{n=1}^\infty f(n)=\sum\limits_{n=1}^\infty 1=\infty$.
Of course you can take triangles of lower height if you want $f(x)<1$
If you want to strengthen your conditions such a function can be made $C^\infty$ too via bump functions
Edit: actually your question is more similar to asking wether if a continuous positive function has a convergent integral then $\lim\limits_{x\to +\infty} f(x)=0$, which is false, as I've shown above, but it's true if you assume $f$ uniformly continuous instead
A: The comments have shown the flaw in the statement, but it looks like ideally you want $q_n=n$ or something similar. That's still possible. You construct $f$ to be zero outside of $(n-1/n^2,n+1/n^2)$, and on those intervals a triangle (i.e., $f(n)=1/2$, linear on the rest of the interval). Then $f(n)=1/2$ for all $n$, and 
$$
\int_0^\infty f=\sum_{n=1}^\infty \frac{1}{2n^2}<\infty.
$$
A: The question as NOW phrased is this:

Does there exist a non-negative continuous function $f:\mathbb{R}_+ \rightarrow \mathbb{R}_+$ which has a convergent integral: $$ \int_0^\infty f \, dx < \infty \tag{1}$$ with the restriction that $\forall x \in \mathbb{R}_+, f(x) <1$. But, which has a strictly increasing sequence $ \{q_n\}_{n=0}^\infty \subset \mathbb{R}_+$ such that: $$\sum_{n=0}^\infty f(q_n)=\infty \tag{2}$$

Certainly $q_n = 1 - \dfrac 1 {2^n}$ satisfies those requirements.
I'm guessing maybe you wanted $q_n\to\infty$ as $n\to\infty$ although you didn't say so.  Taking $\mathbb R_+$ to exclude $0$, how about this:
\begin{align}
q_n & = n \\[10pt]
f(x) & = e^{-x} + g(x) \\[10pt]
\end{align}
where $g(x)=0$ when $x$ is remote from all integers and $g(x)=1$ when $x$ is an integer, and $g(x)$ is nonzero near an integer $n$ only within a distance $2^{-n}$ from $n$. That way the integral of $g(x)$ over an interval centered at $n$ is $\le 2^{-n}$ but $f(q_n)=1$.
