Inequality on integrals of $L^1$ functions

Question

Let $\lambda \geq 0$ and $(X,d,\mu)$ be a $\sigma-$finite measure space. Then for $f, g \in L^1(X,\mu)$ $$ \left| \int_X (|f|-\lambda)^{+} d\mu - \int_X (|g|-\lambda)^{+} d\mu \right| \leq \int_X ||f|-|g|| d\mu$$ holds (where $(x)^{+} = \text{max}(x,0)$). I tried dividing $X$ space into sets where, $(|f|-\lambda)^{+}$ = $(|f|-\lambda)$ and so on, but I still did not manage to prove this inequality. Could you offer me some hints or help with the proof?

Answer 1

Verify the the inequality:

$|x^{+}-y^{+}| \leq |x-y|$ for all real numbers $x$ and $y$.

If you bring the absolute value sign in LHS inside the integral this gives LHS $\leq \int |(|f)-\lambda| -(|g|-\lambda))|=\int ||f|-|g|| =$ RHS.