Let $f$ be a continuous function on $\mathbb{R}$ satisfying $\int_\mathbb{R}|f(x)|dx<\infty$. Can we conclude that $\sum_\mathbb{Z}|f(k)|<\infty$? Let $f$ be a continuous function on $\mathbb{R}$ satisfying
$$\int_\mathbb{R}|f(x)|dx<\infty.$$
Can we conclude that
$$\sum_\mathbb{Z}|f(k)|<\infty?$$
Note: Continuity is necessary otherwise $f=\chi_\mathbb{Z}$ would give counterexample.
 A: No; having finite integral does not impose what $f$ has to behave like on the integers (a set of measure zero). To give a smooth counterexample, we just want to "smooth out" your $\chi_{\mathbb{Z}}$ example. Let $\phi$ be a smooth bump function on $\mathbb{R}$ with compact support in $(-1/2, 1/2)$, with $\phi(0) = 1$, and with $\int_{-1}^1 \phi(x)\, dx = 1$. Define $\phi_n(x) = \phi(n^2(x - n))$. These are narrow versions of $\phi$. They are supporte in the interval $(n - 1/2n^2, n + 1/2n^2)$. Then set
$$
f(x) = \sum_{n \in \mathbb{Z}} \phi_n(x).
$$
Note that for each $x$, the sum is actually finite since the supports of $\phi_n$ are all disjoint.
Then $f(n) = 1$ for each $n \in \mathbb{Z}$ so certainly the sum over the integers isn't finite. However we can compute that
$$
\int_{\mathbb{R}} \phi_n(x) \, dx = n^{-2}\int_{\mathbb{R}}\phi(y) dy = n^{-2}
$$
and thus
$$
\int_{\mathbb{R}} f(x) \, dx = \sum_{n \in \mathbb{Z}}\int_{\mathbb{R}} \phi_n(x)\, dx = \sum_{n \in \mathbb{Z}} \frac{1}{n^2} < \infty.
$$
To get your desired conclusion from the hypotheses, we'd have to suppose some monotonicity on $|f(x)|$, I believe.
A: Let $t(x) = \max(0,1-|x|)$. Then $\int t = 1$ and $t(0) = 1$.
Then let $\phi(t) = \sum_n t(n^2(x-n))$. Then $\int \phi = \sum_n {1 \over n^2}$
but $\phi(n) = 1$ for all $n$, so $\sum_n \phi(n) = \infty$.
Note, with $\eta(t) = \sum_n {1 \over n}t(n(x-n))$ we have $\int \eta = \sum_n {1 \over n^2}$ and $\eta(n) = {1 \over n}$ so
$\sum_n \eta(n) = \infty$ and $\eta \in C_0$.
A: In measure theoretic point of view, consider measure space $(\Omega, \mathcal{F}, \mu)$ and consider a $\mu$-measurable function $f$.
Now consider,
$$
\int f d\mu
$$
In your case you take first $\mu=\lambda$, Lebesgue measure and then in your second case you take $\mu$ as a counting measure, say $\mu'$. But your $\Omega$ is same in both cases.
Now consider $\Omega = [0,1]$ then $\lambda(\Omega)<\infty$ but $\mu'(\Omega)=\infty$. that become problematic, to conclude about summability.
