Prove that $\int_{0}^{+\infty}f(x)dx$ converges. Let $f: [0, +\infty) \to \mathbb{R}$ continuous function and $$\sum_{n\geq1}f(n + x)$$ converges uniformly for $x \in [0, 1]$.
Is there any ability to prove to prove that in this case $$\int_{0}^{+\infty}f(x)dx$$ also converges?
Here's what I think about it:
We only know that $$ \forall \varepsilon > 0 \ \exists N \ \forall n>N \ \forall x \in [0, 1] \ |S_n(x)−f(x)|< \varepsilon $$
where $$S_n(x) = \sum_{k = 1}^{n} f_k(x)$$ and $$f_n(x) = f(n + x)$$
But how can we use this fact for integral convergence?
I also know that $\int_{0}^{+\infty}f(x)dx$ converges iff $\sum_{n\geq1}f(n)$ converges but seems like we don't know anything about $f$.
 A: Yes; first we assume $f\geq 0,$ then deduce the general case.
When $f\geq 0,$ we can write the identity
$$\int_0^{\infty} f(x) \, dx = \sum_{n=0}^{\infty} \int_n^{n+1} f(x) \, dx = \sum_{n=0}^{\infty} \int_0^1 f(x+n) \, dx,$$
valid even when the integral is infinite thanks to non-negativity.
By Tonelli's theorem, for non-negative functions we can interchange summation and integration, to find that
$$\int_0^{\infty} f(x) \, dx = \int_0^1 \sum_{n=0}^{\infty} f(x+n) \, dx.$$
If $f$ is continuous and the partial sums converge uniformly on $[0,1],$ then the infinite sum must be continuous on $[0,1]$ (a uniform limit of the continuous partial sums is continuous). Therefore this integral is finite, as a continuous function integrated over a compact set has finite integral.
Now the general case follows by writing $f = f^+ - f^-$ into a combination of its positive and negative parts, and applying this argument to each of them individually.
A: If by converge you mean integrable in a Lebesgue sense then the answer seems to be no.
Define $b:\mathbb{R} \to \mathbb{R}$ by
$b(x) = \max(0, 1-2|x-{1 \over 2}|)$. Note that $b$ is continuous, the support of $b$ is $[0,1]$ and $b(x) \in [0,1]$.
Let $f(x) = \sum_{k \ge 0} (-1)^{k+1}{1 \over k+1}b(x-k)$. Note that for a given $x \in [0,1]$ there is at most one non zero element in the summation.
It is straightforward to check that $f$ is continuous but not $L^1$.
Let $\phi_N(x) = \sum_{n \ge 1}^N f(x+n)$, $\phi(x) = \lim_N \phi_N(x)$.
Then $\phi(x)-\phi_N(x) = \sum_{n>N} \sum_{k \ge 0} (-1)^{k+1}{1 \over k+1}b(x+n-k) $.
If $x \in [0,1)$ then $\phi(x)-\phi_N(x) = \sum_{n>N}  (-1)^{n+1}{1 \over n+1}b(x) $ (if $x=1$ then $\phi(x)-\phi_N(x) = 0 $) and so
$|\phi(x)-\phi_N(x) | \le |\sum_{n>N}  (-1)^{n+1}{1 \over n+1}|$. In particular, the convergence is uniform.
