This is a good question! It is all a matter of how to define an improper integral. The traditional way is to write
$$\int_a^b f(t)\, dt = \lim_{x \to a} \lim_{y \to b} \int_x^y f(t)\, dt$$
whenever one of the bounds is worrisome. In particular, we can write
$$\int_a^b f(t) \,dt = \lim_{x \to a} \lim_{y \to b} \left(\int_x^1 f(t) \, dt + \int_1^y f(t)\, dt \right)$$
or
$$\lim_{x \to a} \lim_{y \to b} \left(\int_x^1 f(t) \,dt \right) + \lim_{x \to a} \lim_{y \to b} \left( \int_1^y f(t) \,dt \right) $$
so that
$$\int_a^b f(t)\, dt = \lim_{x \to a} \left(\int_x^1 f(t) \,dt \right) + \lim_{y \to b} \left(\int_1^y f(t)\, dt \right).$$
Now convergence means that the integral is finite and the only way for the sum of two numbers to be finite is if both are finite. Hence your professor's comment. All of the above assumes $a < 1 < b$ and that there is no trouble with $f$ near $t=1$.
But you are right that you can get cancelation if you take the limits differently. Instead of taking the limit to a and then the limit to b sequentially, if you took them at the same time--that is to say you combined them--you could possibly get cancellation.
Here's an example. If define instead the improper integral as
$$\int_{-\pi/2}^{\pi/2} \tan t\, dt = \lim_{x \to \pi/2} \int_{-x}^x \tan t\,dt$$
then
$$\int_{-\pi/2}^{\pi/2} \tan t\, dt = \lim_{x \to \pi/2} \left(- \ln \cos(x) + \ln \cos(-x) \right) = 0.$$
What you are thinking of is a useful concept and it often comes up with the term Principal Value in more advanced calculus courses.
No, the equality does not hold in general.
EXAMPLE $1$
For a first example, the integral $I(x)$ as given by
$$I(x)=\int_{-\infty}^\infty\frac{\sin(xt)}{t}\,dt$$
converges uniformly for all $|x|\ge \delta>0$. But the integral of the derivative with respect to $x$, $\int_{-\infty}^\infty \cos(xt)\,dt$ diverges for all $x$.
EXAMPLE $2$
As another example, let $J(x)$ be the integral given by
$$J(x)=\int_0^\infty x^3e^{-x^2t}\,dt$$
Obviously, $J(x)=x$ for all $x$ and hence $J'(x)=1$. However,
$$\int_0^\infty (3x^2-2x^4t)e^{-x^2t}\,dt=\begin{cases}1&,x\ne 0\\\\0&,x=0\end{cases}$$
Thus, formal differentiation under the integral sign leads to an incorrect result for $x=0$ even though all integrals involved are absolutely convergent.
Sufficient Conditions for Differentiating Under the Integral
If $f(x,t)$ and $\frac{\partial f(x,t)}{\partial x}$ are continuous for all $x\in [a,b]$ and $t\in \mathbb{R}$, and if $\int_{-\infty}^\infty f(x,t)\,dt$ converges for some $x_0\in[a,b]$ and $\int_{-\infty}^\infty \frac{\partial f(x,t)}{\partial x}\,dt$ converges uniformly for all $x\in [a,b]$, then
$$\frac{d}{dx}\int_{-\infty}^\infty f(x,t)\,dt=\int_{-\infty}^\infty \frac{\partial f(x,t)}{\partial x}\,dt$$
Best Answer
If $f(x)$ and $g(x)$ are each Riemann integrable on $[0,b]$ for every $b > 0$, then $f(x) + g(x)$ is also Riemann integrable there, and $$ \int_0^b (f(x) + g(x))\; dx = \int_0^b f(x)\; dx + \int_0^b g(x)\; dx$$ By definition,
$$\eqalign{\int_0^\infty (f(x) + g(x))\; dx &= \lim_{b \to \infty} \int_0^b (f(x) + g(x))\; dx\cr &= \lim_{b \to \infty} \left(\int_0^b f(x)\; dx + \int_0^b g(x)\; dx \right)\cr &= \lim_{b \to \infty} \int_0^b f(x)\; dx + \lim_{b \to \infty}\int_0^b g(x)\; dx \cr &= \int_0^\infty f(x)\; dx + \int_0^\infty g(x)\; dx}$$ if those last two improper integrals exist.
However, it can happen that the improper integrals of $f$ and $g$ don't exist, but that of $f+g$ does.