Must $a,b>1$ for $\int_1^a\frac 1tdt+\int_1^b\frac1tdt=\int_1^{ab}\frac1tdt$? In Spivak's Calculus (4th edition), Chapter 13, problem 15, the author asks to prove: 

For $a,b>1$ prove that:
  $$\int_1^a\frac 1tdt+\int_1^b\frac1tdt=\int_1^{ab}\frac1tdt$$ 
  The result is used in some problems in the following chapter, but I can't find in it's proof why $a,b$ must greater than 1, and the problem 14-28(c)(shown in the end) do use it regardless $a,b$'s range.

Here's the proof in answer book:

Notice that
  $$
\frac1b\cdot \inf\{ \frac1t:t_{i-1}\leq t\leq t_i \}=\inf\{\frac1t:bt_{i-1}\leq x\leq bt_i \}
$$
  Denoting the first $\inf$ by $m_i$ and the second by $m_i^\prime$, we have
  \begin{aligned}
L(f,P\prime)
&=\sum_{i=1}^nm_i^\prime(bt_i-bt_{i-1})\\
&=\sum_{i=1}^nbm_i^\prime(t_i-t_{i-1})\\
&=\sum_{i=1}^nm_i(t_i-t_{i-1})\\
&=L(f,P).
\end{aligned}
  So
  $$
\int_b^{ab}\frac1tdt=\sup\{L(f,P\prime)\}=\sup\{L(f,P)\}=\int_1^a\frac1tdt.
$$

Finally, in Chapter 14 problem 28-c, in the answer book, he just say:

Problem 13-15 implies that
  $$\int_{1/2}^1\frac1xdx+...+\int_{1/2}^1\frac1xdx=\int_{1/2^n}^1\frac1xdx$$

even reverse$\int_{1/2}^1$ to $-\int_1^{1/2}$, still doesn't make sense if $a,b$ can not be less than $1$.
 A: If I recall correctly, in chapter 13, Spivak only defined $\int_a^b f$ for $a < b$. So the upper limit is greater than the lower limit. It is only in chapter 14 that he defines the integral $\int_a^bf$ when $b\leq a$. I think this is why he imposes the condition $a,b > 1$ in your question. After defining things appropriately, it should be pretty easy to deduce the equality even when $a,b \leq 1$.

Edit: 
I just checked my copy of Spivak (3rd Edition). He actually defines $\int_a^b$ for $b \leq a$ in chapter 13, right after Theorem 13-4.
In the $3$rd edition, the question reads

Prove that
  \begin{align}
\int_1^a \dfrac{1}{t} \, dt + \int_1^b \dfrac{1}{t} \, dt = \int_1^{ab} \dfrac{1}{t} \, dt
\end{align}
  Hint:   ......

So, he actually doesn't impose any restriction on $a,b$ (the only restriction we should impose is $a,b > 0$). So, actually, you might have to do some case work. If in your edition he says $a,b > 1$, then that's fine, but eventually he'll make use of this property for all $a,b > 0$, so you should see how to extend to proof to all cases.

Additional Remark:
Spivak proves this equality using the Fundamental Theorem of Calculus in Theorem 18-1, for all $a,b > 0$. Even though it is more work, you should still try to see how to extend the "proof by partitions" to all cases. 
A: For $a,b\gt1$:
$$
\begin{align}
\int_1^a\frac1t\mathrm{d}t+\int_1^b\frac1t\mathrm{d}t
&=\int_1^a\frac1t\mathrm{d}t+\int_a^{ab}\frac1{t/a}\mathrm{d}t/a\tag1\\
&=\int_1^a\frac1t\mathrm{d}t+\int_a^{ab}\frac1t\mathrm{d}t\tag2\\
&=\int_1^{ab}\frac1t\mathrm{d}t\tag3
\end{align}
$$
Explanation:
$(1)$: substitute $t\mapsto t/a$ in the right integral
$(2)$: cancel factors
$(3)$: addition of integrals with adjacent intervals

We can do the same thing for $0\lt a,b\lt1$:
$$
\begin{align}
\int_a^1\frac1t\mathrm{d}t+\int_b^1\frac1t\mathrm{d}t
&=\int_{ab}^b\frac1{t/b}\mathrm{d}t/b+\int_b^1\frac1t\mathrm{d}t\tag4\\
&=\int_{ab}^b\frac1t\mathrm{d}t+\int_b^1\frac1t\mathrm{d}t\tag5\\
&=\int_{ab}^1\frac1t\mathrm{d}t\tag6
\end{align}
$$
Explanation:
$(4)$: substitute $t\mapsto t/b$ in the right integral
$(5)$: cancel factors
$(6)$: addition of integrals with adjacent intervals
A: This is the fact $\ln ab=\ln a+\ln b$.  This is true for any $a,b\gt0$.
