unit step signal as sum of impulses from MIT OCW video From MIT's OCW on Signals and Systems by A. Oppenheim, at 4:30 mark https://www.youtube.com/watch?v=jGk3w1b7UXQ&amp;index=3&amp;list=PLLNp7XoiSLQYygYw8Mzt763zZCQdzCZcd

He said that by forming the sum from minus infinity to some value n of a unit impulse function is equal to the unit step function.
$$u[n]=\sum_{m=-\infty}^n \delta[m]$$

At n&lt;0 the sum is accumulating nothing. We see that indeed it is equal to the unit step since the unit step is zero at n&lt;0.

However, I am confused why at n>0, the sum can be equal to the unit step. From the diagram in the video, we see that there is only one impulse value that can be accumulated in the sum. But then the unit step is not composed of just one impulse, rather it is composed of an impulse train. 

The expression is not adding an impulse train since the impulse function is defined only at m=0 which is 1. At m=1,2,3..the impulse function is already zero, so there is just one impulse that is contained in the expression.

What is my wrong assumption/understanding?
yes there is only one term in the sum when $n \ge 0$
$$ \delta[m]=\begin{cases}1 \quad m=0 \\
0 \quad \text{otherwise} \end{cases}$$

$$ \begin{align}\sum_{m=-\infty}^n \delta[m] &amp;= \delta[-\infty]+ \dots +\delta[0] + \dots + \delta[n] \\
&amp; = 0 + \dots + 1 + \dots + 0  \\
&amp;= 1  \; \text{for} \; n\ge 0
\end{align}$$
The sum holds the value of 1 for $n\ge0$. His figure shows the terms inside the sum, not the sum of terms.

The definition of $u[n]$ is
$$ 
u[n]=\begin{cases} 1  \quad \text{for}\; n \ge 0 \\
0  \quad \text{for}\; n&lt;0 \end{cases} $$
They are equivalent.  Defining $u[n]$ function as a sum of delta functions may seem strange but I'm sure he has a "trick" to show that will be useful down the road.

There is a similar relationship in continuous time relating the unit step function $u(t)$ as the integral of the Dirac $\delta(t)$.