$\forall x\in [0,1]$, if $f(t)f(t-x)$ is integrable, then $f(t)$ is integrable? The original problem (from here problem 8) is: 
Let $\alpha>\frac{1}{2}$ be a real number. Prove that it is impossible to find a real function $f$ such that
$$f(x)=1+\alpha\int_x^1 f(t)f(t-x) \, dt$$
holds for all $0\leq x\leq1$.
My solution: Integrate both sides for $x$ from $0$ to $1$ and suppose $s=\int_0^1f(x) \, dx$. Then the equality becomes $s=1+\frac{1}{2}\alpha s^2$. Apply AM-GM to finish the proof.
However, my solution works only when $f(x)$ is integrable (on $[0,1]$). 
My question is: if the equality in the problem holds, then $\forall x\in [0,1]$, $f(t)f(t-x)$ is integrable, so is it true that $f(x)$ is integrable? If it is not true, is there an explicit counterexample?
For all "integrable" here, I think I mean "Riemann integrable". I have not studied the Lebesgue theory yet.
 A: Suppose the equality holds for all $x$. In particular at $x=0$ we notice that
$$\|f\|^2_{L^2[0,1]} = \int_0^1 |f(x)|^2 \ dx  < \infty.$$
Then by Cauchy-Schwarz for functions,
$$ \int_0^1 |f(x)|\ dx = \langle |f|, 1\rangle_{L^2[0,1]}\leq \|f\|_{L^2[0,1]}\|1\|_{L^2[0,1]} = \|f\|_{L^2[0,1]} < \infty.$$
Thus, as long as $f$ is (Lebesgue) measurable, your proof works.
A: Almost a counterexample, The example below is a counterexample excepting for $x=0,1$ :(
Maybe someone smarter can modify it .
Let $f:[0,1] \to \mathbb R$.
$$
f(t)=
\left\{ \begin{array}{lc}
1 & \mbox{ if } t =\frac{n}{p}  \mbox{ with } n \in \mathbb N \mbox{ and } p \mbox{ prime } \\
-1& \mbox{ otherwise }
\end{array}
\right.$$
Now, since the set $A=\{ t= \frac{n}{p} :  n \in \mathbb N \mbox{ and } p \mbox{ prime }  \}$ is dense in $[0,1]$ it follows that $f$ is discontinuous everywhere and hence not Riemann integrable.
Claim: For all $x \in (0,1)$ the function $f(t)f(t-x)$ is Riemann integrable.
Proof:
Case 1: If $x \notin \mathbb Q$ then it is easy to see that $f(t)f(t-x)=0 \forall t \in [0,1]$, which is Riemann Integrable.
Case 2: If $x =\frac{m}{n} \in \mathbb Q$ then it is easy to see that $f(t)f(t-x)=0 \forall t \in [0,1]$.
We claim that in this case, for each such fixed $x$, we can only find finitely many $t \in A$ such that $t+x \in A$.
So let $t= \frac{k}{p}$ be so that $t, t-x \in A$.
We show next that $p|n$. 
Assume by contradiction that $p \nmid n$ then 
$$t-x=\frac{k}{p}-\frac{m}{n}= \frac{kn-pm}{np}$$
Now, since $p|pm$ and $p \nmid kn$, $p$ cannot be reduced from the second fraction. 
Since the second fraction must belong to $A$, and hence must be reduced to a fraction to prime denominator, it follows that $n$ must be reduced and hence 
$$n|kn-pm \Rightarrow n|pm \Rightarrow n|m$$
with the last implication following from $p$ prime and $p \nmid n$ implies $gcd(p,m)=1$.
Therefore $n|m$ and hence $x=\frac{m}{n} \in \mathbb Z$ contradiction.
This yields that $p|n$ and hence 
$$t \in \{ \frac{k}{p} : p |n, 1 \leq k \leq p \}=:A_n$$
Now, since $x$ is fixed, so is $n$, and each fixed $n$ has finitely many prime divisors, from where it follows that $A_n$ is finite.
To complete the proof of Case 2:
Since there are only finitely many such $t$, it follows that $f(t)f(t-x)=0$ outside this finite set. Therefore, $f(t)f(t-x)$ is Riemann integrable.
