Proof that if the MRS is increasing in one good, then the other is normal

Question

Osborne and Rubinstein (Models in Microeconomic Theory, p. 70) prove the proposition that, if the demand function of a rational consumer has the property that $\text{MRS}(x_1, x_2)$ is increasing in $x_2$ for all values of $x_1$, then good 1 is normal, as follows:

Fix $p_1^0$ and $p_2^0$ and let $w' > w$. Let $a$ be a solution to the consumer's problem for the budget set $B((p_1^0, p_2^0), w)$ and let $a'$ be a bundle on the frontier of $B((p_1^0, p_2^0), w')$ with $a_1' = a_1$. By the assumption on the marginal rate of substitution we have $\text{MRS}(a') > \text{MRS}(a)$, and hence the solution $b$ of the consumer's problem for the budget set $B((p_1^0, p_2^0), w')$ has $b_1 > a'_1 = a_1.$

I don't understand the very last inference. Why does the fact that $\text{MRS}(a') > \text{MRS}(a)$ imply the desired conclusion?

Answer 1

Note that $(a_1,a_2)$ is the utility maximizing bundle at relative price $MRS(a) = (= p_1/p_2)$ and budget $w = MRS(a).a_1 + a_2$.

1. First as $b$ is utility maximising, $MRS(b) = \frac{p_1^0}{p_2^0}$. As $MRS(a') > MRS(b) = MRS(a)$ and the $MRS$ is unique (if utility is differentiable) we have that $a_1' \ne b_1$.

2. Now for the proof. Assume, towards a contradiction, that $b_1 \le a_1'$, which gives by point 1 above that $b_1 < a_1$. Then: $$ \begin{align*} MRS(a') (a_1' - b_1 ) + a_2' - b_2 &= MRS(a') (a_1' - b_1) + w - \frac{p_1^0}{p_2^0} a_1' - w + \frac{p_1^0}{p_2^0} b_1,\\ &= \underbrace{(MRS(a') - MRS(a))}_{> 0} \underbrace{(a_1' - b_1)}_{> 0} > 0 \end{align*} $$ As such, we get that: $$ MRS(a') a_1' + a_2' > MRS(a') b_1 + b_2. $$ As $(a_1', a_2')$ is utility maximising for relative prices $MRS(a')$ and budget $w'' = MRS(a') a_1' + a_2'$ and $b$ was cheaper to buy at that price and budget, we obtain by a revealed preference argument, that $u(a_1', a_2') > u(b_1, b_2)$, contradicting the assumption that $(b_1, b_2)$ was utility maximising at the budget $B((p_1^0,p_2^0), w)$ and that $a' \in B((p_1^0, p_2^0), w)$.