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 $$\nu(E)= \int_E fdm \geq \delta.$$