prove that $ \int_E f(x) dm \ge \delta$ whenever $m(E) \ge \epsilon$

Question

Assume that $f: [0,1] \to [0, \infty) $ is a Lebesgue measurable function such that $ f(x)\gt 0 $ a.e x. Show that for every $ \epsilon \gt 0 $ there is $\delta \gt 0$ such that for every lebesgue measurable E with lebesgue measure $m(E) \ge \epsilon $ we have $ \int_E f(x) dm \ge \delta$.

I have seen similar problem related to $ \epsilon $ and $\delta $ when f is lebesgue integrable. but the inclusion was opposite. I think this should not be too tough but I could not get after spending long time.

Answer 1

Hint: Suppose not, then:

$$\left (\exists \varepsilon >0 \right ) \left ( \forall n \right ) \left ( \exists E_n \subset [0,1] \right ) \left ( \lambda(E_n)\ge \varepsilon \ \mbox{and} \ \int_{E_n}f d\lambda < \frac{1}{n} \right )$$

The fact that you're working with finite measure is important.

Also, since $f>0$ a.e., we know that $\int_A f d\lambda>0$ for any set $A$ of positive measure.

Answer 2

Put $A_k = \{x : f(x) < 1/k\}$ then $\lim_{k\to \infty} m(A_k) = 0$. For any $\epsilon >0$, choose $k$ depending only on $\epsilon$ such that $m(A_k) < \epsilon/2$. By Cauchy-Schwartz we have $$m(E\cap A_k^c) \leq \left(\int_{E\cap A_k^c} f dm\right)^{1/2} \left(\int_{E\cap A_k^c} \frac 1f dm\right)^{1/2} \leq m(E)^{1/2} \left(k \int_{E} f dm\right)^{1/2}.$$ Using inequality $ab \leq a^2 + b^2/4$, we have $$m(E\cap A_k^c) \leq k\int_{E} f dm + \frac{m(E)}4.$$ Hence $$m(E) = m(E\cap A_k^c) + m(E\cap A_k) \leq k\int_{E} f dm + \frac{m(E)}4 + m(A_k) \leq k\int_{E} f dm + \frac{m(E)}4 +\frac\epsilon 2, $$ or $$\frac{3m(E)-2\epsilon}4 \leq k\int_{E} f dm.$$ Then, if $m(E) \geq \epsilon$ we obtain $$\int_E f dm \geq \frac{\epsilon}{ 4k}.$$

Answer 3

As you say, we have all seen something quite similar but with an opposite inequality. That rather suggests they are closely related and perhaps one can be deduced from the other. Or not.

In this case maybe and it is a bit entertaining so I offer a sketch here. When things look vaguely familiar (even if upside down) it is very tempting either to look for similar methods or, even more tempting, to deduce one version from the other.

Define a measure $\nu(E) = \int_E fdm$ for every measurable subset $E$ of $[0,1]$. (Not necessarily finite, but certainly $\sigma$-finite.) Define $g(x) = 1/f(x)$ for every $x$ then $g$ is measurable, $f(x)g(x)=1 $ a.e., and $g$ is almost everywhere finite. Moreover $g$ is $\nu$-integrable on $[0,1]$ since for each measurable set $A$ $$\int_A g\, d\nu = \int_ A fg\, dm = m(A) \leq 1.$$

Now we apply the usual version in this way. Let $\epsilon>0$ and choose $\delta>0$ so that if $\nu(A) <\delta$ then $\int_ A g\,d\nu < \epsilon$.

Now observe that if $$m(E)=\int_ E gd\nu \geq \epsilon$$ this must mean that $$\nu(E)= \int_E fdm \geq \delta.$$