Young's Inequality for Complementary Functions 
I was trying to understand this theorem please see in the image. However I have no idea about the equality which I  framed with red. 
Second, will not Young inequality turn into equality when both $y\geq \varphi(u)$ and $x\geq \psi(v)$? I couldn't understand why equality holds when $x\geq \psi(y)$ and $y=\varphi(x)$ and the relation of $y\geq \varphi(u)$ with $x=\psi(v)$.
P.S. : $(13)$ is the definition of $\psi$ i.e. $\psi(u)=\inf\{t : \varphi(t)>u\}$ where $u\geq 0$ and $\varphi : \mathbb R^+\to \bar{\mathbb R}^+$ is nondecreasing and left continuous creator of the young function $\Phi$.
Thanks in advance for any help. Please forgive me if there are meaningless things.
 A: The equalities follow from the observation that for all $\varepsilon >0,$
\begin{align}
  \varphi(\psi(u)) &\leq u \leq \varphi(\psi(u)+\varepsilon), \\
  \psi(\varphi(v)) &\leq v \leq \psi(\varphi(v)+\varepsilon),
\end{align}
which follows from the definition of $\psi$ and the properties of $\varphi$ (notably left-continuity). Using these we obtain the chain of implications
\begin{align}
  \varphi(v) < u &\implies v \leq \psi(u) \implies \varphi(v) \leq u, \\
  \psi(u) < v &\implies u \leq \varphi(v) \implies \psi(u) \leq v.
\end{align}
Hence up to a negligible set where $\{ u = \varphi(v), v = \psi(u)\},$ the two cases are mutually exclusive.. This explains the first equality, where we split the region $\{ 0 \leq u \leq x, 0 \leq v \leq y\}$ into these two regions where the two cases occur.
The second equality follows along similar lines; what's slightly confusing is that they swap the $u$ and $v,$ but the idea is to only write it in terms of $0 \leq u \leq \varphi(v)$ and $0 \leq v \leq \psi(u)$ respectively. Without swapping the variables this gives
$$ uv = \int_0^x y \wedge \varphi(v) \,\mathrm{d} v + \int_0^y x \wedge \psi(u) \,\mathrm{d} v, $$
as required.
Finally for equality to occur, we require the sets
\begin{align}
  \{ 0 \leq v \leq x : \varphi(v) \geq y\}, \\
  \{ 0 \leq u \leq y : \psi(u) \geq x\},
\end{align}
to be negligible. For the first note that $\psi(y) < v$ implies $\varphi(v) \geq y,$ so this cannot hold for all $v < x.$ Thus $\psi(y) \geq x,$ and arguing similarly for the second gives $\varphi(x) \geq y.$ Combining these we deduce that $\varphi(x)=y$ and $\psi(y)=x.$
