Problem #11 in Royden-Fitzpatrick $4^{th}$ edition. The problem is given below:



EDIT:(According to the hint I received from @Marios Gretsas)
FIRST: Proving this change of variables formula for indicator functions of sets
Since indicator functions of sets $\chi$ are measurable (I know how to prove this) and since Measure is translation invariant by pg.30 in Royden and Fitzpatrick, then the statement is true for indicator functions of sets.
SECOND: Considering simple functions.
Now define a simple function $\psi$ as follows: $$\psi = \sum_{i = 1}^{n} a_{i} . \chi_{E_{i}}$$
Where each $E_{i} = \psi^{-1}(a_{i}) = \{ x \in E | \psi(x) = a_{i}\}$ and where $E_{i}$ are disjoint and $a_{i}$ are not necessarily distinct. 
Now, since the integration given in the question is over the interval $[\alpha + \gamma, \beta + \gamma]$ which is of finite measure, which is the length of the interval. And by the definition of integrations for simple functions given on pg.71, we have $$\int _{[\alpha + \gamma, \beta + \gamma]} \psi = \sum_{i = 1}^{n} a_{i} . m(E_{i})$$
Now, I am stuck, what is the relation between $E_{i}'s$ and my indicator functions?
I am also starting to follow the suggestion of @amsmath  
EDIT:
Also I have a question Do I have to do the following:
reduce from simple functions to characteristic functions of measurable sets, to char. fcns of $G_{\delta}$ sets, to char. fcns. of open sets, to char. fcns. of open intervals.    
EDIT:
I also found this question, I feel it may be helpful. Integration by substitution for Lebesgue integration 
 A: Okay we know what for a simple function $\phi (x) = \sum_{i=1}^{n} a_i\chi_{ E_i} $ defined such that $\bigcup E_i \subset [\alpha+\gamma , \beta+\gamma]$, we have  $$\int_{[\alpha+\gamma,\beta+\gamma]} \phi = \sum_{i=1}^{n} a_i m(E_i)$$. Now 
$$\phi(t+\gamma) = \left\{ \begin{array}{cc} a_1 &, t+\gamma \in E_1 \\ a_2 &, t+\gamma \in E_2 \\ \vdots & \end{array}\right.$$ Now we can define $E'_i = E_i - \gamma $  then  $\bigcup E'_i \subset [\alpha , \beta] $.
$$\phi(t+\gamma) =\phi' (t) = \left\{ \begin{array}{cc} a_1 &, t \in E'_1 \\ a_2 &, t \in E'_2 \\ \vdots & \end{array}\right.$$. Thus finally 
$$\int_{[\alpha , \beta] }\phi(t+\gamma) =\int_{[\alpha , \beta ]} \phi' (t) = \sum a_i m(E'_i) = \sum a_i m(E_i ) = \int_{[\alpha + \gamma, \beta+\gamma]} \phi(x) dx $$. 
Now we know that for bounded function $g$ defined on $E= [\alpha+\gamma ,\beta+\gamma ]$ we define the upper integral of $g$ as 
$$\inf \left\{ \int_E \psi : \psi \text{ is simple and } \psi \geq g \right\}$$ 
and the lower integral as $$\sup \left\{ \int_E \phi : \phi \text{ is simple and } \phi \leq g \right\} $$
And $g$ is integrable if both values are equal. 
and we know from the simple approximation lemma that for a bounded function $g$ on $[\alpha+\gamma  , \beta+\gamma ]$ there exists two simple functions $\phi \leq g \leq \psi $ for any given $\epsilon$ such that $\psi - \phi < \epsilon $. 
Hence given $\epsilon > 0$ there exists $\phi , \psi $ simple functions such that $\phi \leq g \leq \psi $ and $\psi - \phi < \frac{\epsilon}{\beta-\alpha}$ then we have $g \leq \psi $ on $[\alpha + \gamma , \beta + \gamma] $ , then $g(t+\gamma) \leq  \psi (t+ \gamma)=\psi'(t) $ on $[\alpha , \beta]$.   $$\left|\int_{[\alpha , \beta ] } g(t+\gamma) - \int_{[\alpha + \gamma, \beta + \gamma ]}  g(x) \right| \leq \left| \int_{[\alpha , \beta ]} \psi' - \int_{[\alpha + \gamma , \beta + \gamma ]} \phi \right|\leq \int_{[\beta,\alpha]} | \psi - \phi | < \epsilon $$
