Why do we essentially need complete measure space? While reading the motivation of complete measure space on Wikipedia, what I concluded was, 
completeness is not really necessary when we define on one measure space  and it is necessary when we want to measure on product of measure spaces (is it true ?).  If $\lambda$ is measure on $X$ and $Y$ then is it true that $\lambda^2$ is measure of $A$x$B$ and how ? 
I am not able to understand that $\lambda^2(A\times B)=\lambda(A)\times\lambda(B)$ ? 
Essentially what is the flaw in the measure without being complete ? 
Waiting for response. Thanks! 
 A: I am giving an answer of this question Philosophically...
All intuition of Measure Theory comes from Probability Theory (finite measure theory ). In Probability certain event is impossible then all its sub events are also impossible (usually).
A: One place in probability theory where complete measures are used is the theory of stochastic processes.  We have a stochastic process $X_t$ indexed by reals $t$, so there are uncountably many of them.  Certain combinations or these are important, but (as far as can be proved) only equal almost everywhere to a countable combination.  With complete sigma-algebra, that is enough for us to conclude that this combination is measurable.
A: We want measure spaces to be complete because we want to treat sets of measure zero as negligible. For example, if two functions $f$ and $g$ satisfy $f(x)=g(x)$ for all $x\in X\setminus N$, and $N$ has measure zero, then we'd like to treat $f$ and $g$ as essentially the same thing. However, without completeness it's possible that $f$ is measurable but $g$ is not. 
The issue of completeness is brought into light by the product operation, because the product of complete measures is not always complete. For example, let $A\in [0,1]$ be a nonmeasurable set. The set $A\times \{0\}\subset [0,1]\times [0,1]$ is not measurable with respect to the product measure $\lambda\otimes\lambda$. However, $A\times \{0\}\subset [0,1]\times \{0\}$ and the latter set has product measure $0$. So, once we take the completion of the product measure, $A\times \{0\}$ becomes a measure $0$ set.
A: When proving that something holds almost surely, i.e, that its negation never holds (say, the event $A$), it comes handy to 'bound' such an event with a bigger one of measure zero. Then, we do not need to prove that the original event is measurable. 
Now, I am sure we can also think of 'practical' cases in which is hard (or not possible) to prove that $A$ is measurable when there is no completeness, while being easy to bound $A$ it by a set of measure zero.
A: I've spent some time on checking whether the completeness of measures plays a role in theorems. At first glance I couldn't find any advantage, so I think some of you maybe interested in it. 
There is a so-called projection theorem (Measurable projection theorem proof reference), where completeness is essentially needed.
The second theorem where we assume that the measure is complete (it may be relaxed) is the so-called Scorza-Dragoni theorem.
(e.g. https://www.google.pl/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0ahUKEwiAocr7qfDYAhWEjSwKHbdhAC4QFggmMAA&url=http%3A%2F%2Fmatwbn.icm.edu.pl%2Fksiazki%2Ffm%2Ffm138%2Ffm138118.pdf&usg=AOvVaw3fBgcTjGSPTuckQ4HysisV)
In fact, the Scorza Dragoni theorem uses projection theorem.
