A question about a stronger version of Hölder inequality 
Let $\lambda$ be the ordinary Lebesgue measure onto $(0,1)$. Let $a>1$, $f,g \colon (0,1) \to (0,+\infty)$ two $\lambda$-measurable functions and $B \subset (0,1)$ a $\lambda$-measurable subset s.t. 
  $$
a\int_C fg d\mu \le \left( \int_C f^2 d\mu \right)^{\frac{1}{2}}\left( \int_C g^2 d\mu \right)^{\frac{1}{2}}
$$
  for every $C\subset B$ measurable.
  Show that $\lambda(B)=0$.

I think that the problem is interesting but, unfortunately, I have no idea to begin. I've only noticed that the given inequality is a stronger version of Hölder inequality. What do you suggest me? 
Furthermore, I am wondering if the claim is true only for Lebesgue measure or it is true for any - say - probability measure on Borel sets in $(0,1)$. What do you think? 
Thanks a lot. 
 A: My answer borrows a little bit from J.J. but is also interesting. Let $x \in (0,1)$ be common Lebesgue point of $f^2, g^2$ and $fg$. Then we have by assumption for $h > 0$
$$a \, \int_{x-h}^{x+h} f \, g \, d x \le \left(\int_{x-h}^{x+h} f^2 \, d x \right)^{1/2} \left(\int_{x-h}^{x+h} g^2 \, d x \right)^{1/2}.$$
Dividing by $2h$ and passing to the limit $h \to 0$, we obtain the contradiction
$$a \, f(x) \, g(x) \le f(x) \, g(x).$$
In particular, you do not need a sophisticated choice of $C$, but take a rather arbitrary sequence.
A: Assume that B had positive measure. The idea is to find a subset C in B such that f and g are almost constant there, in which case the inequality breaks.
Indeed, let $\varepsilon > 0$. Then there exists a set $C' \subset B$ of positive measure and a constant $b > \delta$ such that $f(x) \in (b,b+\varepsilon)$ for all $x \in C'$. Here $\delta > 0$ is a constant such that $f^{-1}(\delta,\infty)$ and $g^{-1}(\delta,\infty)$ have positive measure. Similarly there exists a set $C \subset C'$ of positive measure and a constant $c > \delta$ such that $g(x) \in (c,c+\varepsilon)$ for all $x \in C$.
Now notice that
$$\left(\int_C f^2 \, d\mu\right)^{1/2} \left(\int_C g^2 \, d\mu\right)^{1/2} \le (b+\varepsilon)(c + \varepsilon)\mu(C)$$
and
$$a \int_C fg \, d\mu \ge abc \mu(C).$$
By choosing $\varepsilon$ small enough, we find that $(b+\varepsilon)(c + \varepsilon) \ge abc$, which is a contradiction, since $\varepsilon > 0$ is arbitrary, and while $b$ and $c$ can depend on $\varepsilon$, they are bounded from below.
