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This is from the Real Analysis by Folland:

Let $\nu$ be a signed measure, then we can write the following: $\nu(E)= \int_{E}fd|\nu|$ where $f=\chi_P -\chi_N$ such that $P \cup N$ is a Hahn Decomposition for $X$.

Now, my question is on the LHS we have some constant(ignoring the case that $\nu(E)=+\infty$ or -$\infty$ for now) but on the RHS we have some function $f$ which includes characteristic functions. So how can it be possible, what is the point that I am missing?

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    On the RHS you don't have a function $f$, but an integral of function $f$ with respect to measure $|\nu|$ (over set $E$). – 2017-01-04

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But the function is integrated over, hence also gives a real number - a "constant". There isn't any variable on the right hand side of your equation.

To the equality itself: Recall that for any measure we have by definition of the integral $$ \nu(E) = \int_E 1_X\, d\nu $$ Now, we have, by definition of $|\nu|$, that $$ |\nu| = 1_P\nu - 1_N\nu \iff \nu = 1_P|\nu| - 1_N|\nu| $$ hence, $$ \nu(E) = \int_E (1_P - 1_N)\,d|\nu| $$

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    $\nu^{+} = 1_P \nu$ and $\nu^{-}=1_N \nu$, right? – 2017-01-04
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    Almost, in the notation I'm used to $\nu^- = -1_N\nu$, so that $\nu^\pm$ are (non-negative) measures. – 2017-01-04
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    I still can not understand the use of char. function. Can we say that $1_P \nu$ means that $ \nu(E \cap P)$ ? – 2017-01-04
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    The measure(!) $1_P\nu$ is the map $$(1_P\nu)(E) = \int_E 1_P \, d\nu = \nu(E \cap P), $$ correct. – 2017-01-04