Let's start with the $L_{2}[0,1]$ norm.
Start with a single function $f$. The 2-norm of $f$ is
$ \| f \|_{2}=\sqrt{\int_{0}^{1} f(x)^2 dx} $
Now, $\| f \|_2$ will be 0 only if the integral of $f(x)^2$ is 0. This will certainly happen if $f(x)=0$, but it can also happen in cases where $f$ has some isolated points with discontinuities where the function values are non zero. Thus the "null vector" in this space is the zero function and all of the functions that are zero "almost everywhere" are considered to be equivalent to $f(x)=0$ with respect to this norm.
If you're familiar with electrical engineering, where $f(x)$ might be a time varying voltage, then $\| f \|_2$ is essentially the energy associated with that signal.
Now, suppose that $f(x)$ and $g(x)$ are two functions defined on $[0,1]$, and consider the 2-norm of the difference:
$ \| f-g \|_{2}=\sqrt{\int_{0}^{1} (f(x)-g(x))^2 dx} $
If the functions $f$ and $g$ are very nearly identical on $[0,1]$, then this norm will be close to 0.
If you're familiar with electrical engineering terminology, you
might have heard of the "root mean square (RMS)" difference between two signals. This norm is exactly the RMS difference between $f$ and $g$.
If $f(x)=g(x)$, except for perhaps some individual points where there are discontinuities, then the norm will actually be 0. Thus the $L_{2}$ distance between two functions is 0 if the functions are equal "almost everywhere."