Complementary slackness conditions (Kuhn-Tucker)

Your intuition is correct. Say you know that $Z=Xcdot Y=0$ You don't know if $X=0$ or $Y=0$ or both are zero. Even if you know that $X=0$ you have no idea if $Y=0$, $Y<0$ , or $Y>0$.

Consider the potentially satiated utility function: $$ max_{X,Y} U(X,Y) = min(X+Y, 5)$$ $$ S.T. :p_x X + p_y Y + p_z Zleq M$$ Assume, for simplicity, that $p_x = p_y = p_z =1$. In lagrangian form this is: $$ max_{X,Y} U(X,Y) = min(X+Y, 5) - lambda (X+Y+Z-M) $$ Z is the free disposal good, in that it uses up extra money but provides no utility. If $M>5$ then the budget constraint binds. Under this condition, $lambda$ is the shadow value of more income, and is also zero.

Or, if that utility function doesn't suit you, consider: $$ max_{X,Y} U(X,Y) = -(X+Y-5)^2 - lambda (X+Y+Z-M) $$
If $X+Y>5$ then the household wants to use free disposal and set $X+Y=5$. The budget constraint doesn't bind and the MU of income is zero: $MU_{X+Y+Z=5}=-2(X+Y-5)=0$.

You are right. The second statement is logically incorrect. To make the point, let me write for convenience sake $tilde{g}(x):=g(x) - b$. Then by the complementary slackness condition, we have

$$lambda cdot tilde{g}(x) = 0 $$

which comes from the Kuhn-Tucker optimality conditions $tilde{g}(x) le 0$ (primal feasibility of the solution) and $lambda ge 0$ (dual feasibility of the solution). By these constraints, we realize that both can hold as equalities, but not as inequalities. However, if $lambda > 0$, then $tilde{g}(x) =0$. This statement is equivalent to the contra-positive statement that if $tilde{g}(x) <0$, then $lambda = 0$. We observe that we can infer from an inequality constraint that the other constraint must hold by equality. However, we cannot infer that if a constraint holds by equality, then the other must be an inequality. This is a fallacy.

It is possible to have

$$g(x^*) = b; {rm and}; lambda^* = 0$$.

When the multiplier is zero and the constraint is equal to zero, then

a) The constraint does not really "bind"

b) That's why the multiplier is zero.

What does it mean "the constraint does not really bind"?

It means that the solution $x^*$, that makes $g(x^*) = b$, would be chosen even if the constraint was not imposed. In that sense, the constraint is not really binding, because it does not really forbid us to go where we wanted to be, because we are already there.

Consider the simple example

$$max_x {-ax^2 + bx},qquad s.t. ;x geq frac{b}{2a} $$

The Lagrangean is

$$Lambda = -ax^2 + bx + lambda left(x -frac{b}{2a}right)$$

and the f.o.c is

$$x = frac{b+lambda}{2a}$$.

Try cases:

a) $lambda^* = 0$ leads to $x^* = frac{b}{2a}$, which is the unconstrained f.o.c. also.

b) $lambda^* > 0$ the f.o.c. initially indicates that $x^* > b/2a$. But then the constraint is not binding and we should have $lambda^* =0$ : contradiction.

So we see that in this case the solution is

$$x^* = frac{b}{2a},;;; lambda^* = 0.$$

So the constraint appears to be binding, but it really is not.