Why if $m(\varphi(A))\leq \int_A|\det \varphi'|$ doesn't hold, there is $\varepsilon>0$ s.t. $m(\varphi(A))>\int_A|\det \varphi'|+\varepsilon m(A)$?

Question

Let $A$ a closed cube, $m$ the Lebesgue measure on $\mathbb R^n$ and $\varphi:U\to \mathbb R^n$ a $\mathcal C^1$ application (but this point is not important for my question). I recal that $\varphi'$ is the Jacobian matrix. Why if $$m(\varphi(A))\leq \int_A|\det \varphi'|$$ doesn't hold for a closed cube of side $a$, then there is $\varepsilon>0$ s.t. $$m(\varphi(A))>\int_A|\det \varphi'|+\varepsilon m(A)\ \ ?$$

For me it would be only : there is a cube of side $a$ s.t. $$m(\varphi(A))>\int_A |\det \varphi'|,$$ I don't understand the thing with the term $\varepsilon m(A)$.

Answer 1

Hint

$$a\leq b\iff \forall \varepsilon>0, \ \ a<b+\varepsilon \iff \exists A\subset \mathbb R^N : \forall \varepsilon>0,\ \ a <b+\varepsilon m(A).$$