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

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?

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 $$a' \in B((p_1^0, p_2^0), w)$$.

• Thank you for the reply. Sorry, I still have some doubts: why are we allowed to let $\text{MRS}(b) = \text{MRS}(a) = p_1^0/p_2^0$ without assuming anything about the consumer's utility? If the optimal choice of either good 1 or good 2 is zero, the marginal rate of substitution need not be equal to the price ratio. Also, why is $(a_1', a_2')$ utility-maximising? This is not true neither by assumption nor by construction, no? Commented Jun 24 at 16:11
• Plus, I don't understand the revealed-preference argument you made. Can't a contradiction be equally derived from the result that $\text{MRS}(a')a'_1 + a'_2 > \text{MRS}(b)b_1 + b_2$ by noting that by assumption $a'$ was on the frontier of $B((p_1^0, p_2^0), w')$ and $b$ was optimal (so by monotonicity also on the frontier), and so we would have $w' > w'$ which is clearly a contradiction? This would require $\text{MRS}(a')a'_1 + a'_2 = w'$ though, which relates to my previous question about how you know that $a'$ is optimal. Commented Jun 25 at 12:08
• @Riccardo Iorio For corner solutions, you need to be careful as as derivatives (of the utility function) are not defined on the boundary of the domain. As such, there is no MRS. The hyperplane through the (utility maximising) boundary point with slope $(p_1^0/p_2^0)$ is still tangent to the upper contour set. So one needs convex analysis to analyse such cases.
– tdm
Commented Jun 25 at 16:00
• Revealed preference tells that if one bundle was chosen and a second bundle was also affordable for that budget but cheaper to buy, then the chosen bundle should give a higher utility than the second bundle.
– tdm
Commented Jun 25 at 16:07
• Thanks again. Can you just clarify why $a'$ is optimal? By construction it's on the boundary of $w'$ but is not optimal ($b$ is). Commented Jun 25 at 16:15