Given a pmf, how is it possible to calculate the cdf? Given a pmf (probability mass function) for X (random variable): 
\begin{array}{|c|c|c|c|c|}\hline x&1&2&3&4\\ \hline p(x)&0.4&0.3&0.2&0.1\\ \hline \end{array}
How would you know the cdf which is for values for $1 \leq x < 2$ since you do not know the values in between?
 A: Given that you're talking about a discrete random variable over the integers, you should certainly know how the pmf behaves between those values - the probability of it taking any value in any interval strictly between (say) $1$ and $2$ is $0$.
Consequently, you do also know how the cdf $F(x)$ behaves, since it's just the sum of all the probabilities up to $x$. It doesn't matter how many zeroes you want add in, they don't change anything.
For further discussion of this point, see Wikipedia's Cumulative Distribution Function; Definition
... and the two sections immediately under that (Properties and Examples). You may find the drawing of a discrete cdf at the right hand side of the Properties section helpful (it's the top one).
Here's an example for a slightly different distribution than the one in your question (though it's broadly similar).

A: You can compute the CDF using delta-functions. Express the PMF as follows, 
$$ p(x) = (0.4) \delta(x-1) + (0.3) \delta(x-2) + (0.2) \delta(x-3) + (0.1) \delta(x-4) $$
The CDF is then given by integration, by definition, if $P(x)$ is the CDF then, 
$$ P(x) = \int_{-\infty}^x p(y) ~ dy $$ 
Observe that if $x<1$ then each of the delta functions vanish and so $P(x) = 0$. 
If $1<x<2$ then the only delta function which contributes to the integral is $\delta(x-1)$, so we see that $P(x) = (0.4)$ on this interval. 
The same procedure can now be carried out on the other invervals. 
