Property of Lebesgue Integration Let $f$ be bounded measurable function on a set of finite measure $E$. For a measurable subset $A$ of $E$, I want to show that $\int_Af=\int_Ef.\chi_A$ .
I only know that I have to prove both the inequalities. But I just don't know how to start and which other property to use and how? Because it looks just so trivial!
Thank you for the help.
 A: This probably is not right answer for your question, because your and "mine" definition of Lebesgue integral are different. But anyway, I will show how we can see this if we are using definition like in my book.
Before starting, just to say that we are working on some measurable space $(X,\mathfrak{M},\mu)$, where $\mu$ is some given measure on $\sigma$ algebra $\mathfrak{M}$ on set $X$.

Definition 1. If $s=\sum_{k=1}^{n}c_k \chi_{E_k}$ is standard representation of positive simple function $s$ on $X$, $A \subset X$ measurable set, then we define integral function $s$ on set $A$ like $$\int_A s d\mu = \int_A s(x)d\mu(x) = \sum_{k=1}^{n} c_k \mu(A \cap E_k)$$

We can "easily" that for every simple positive function $s$ and for every measurable set $A \subset X$ its true $$\int_A s d\mu = \int_X s \chi_A d\mu.$$ 
Proof I will leave for you, just use definition for integral of simple positive function $s=\sum_{k=1}^{n}c_k \chi_{E_k}$ and fact that $s\chi_A=\sum_{k=1}^{n}c_k \chi_{A\cap E_k}$.
Now, we can define integral of positive function. Let $f:X \to [0,+\infty]$ be measurable function. With $\Gamma_f$ we denote set of all simple positive function $s$ on $X$ such that $s(x) \le f(x)$ for all $x \in X$. Then we define 

Definition 2. Let $f:X \to [0,+\infty]$ is measurable function on $X$ and let $A$ is measurable subset of $X$. Then we define integral function $f$ over set $A$ like $$\int_A f = \sup_{s \in \Gamma_f} \int_A s d\mu.$$

We can see that if $f$ is simple function, then $\displaystyle\int_A f d\mu = \max_{s \in \Gamma_f} \int_A s d\mu,$ so our definition of integral for positive function naturally extends the definition 1 for simple positive function.
Next lemma is very important for us:

Lemma. If $f:X \to [0,+\infty]$ is measurable function on $X$ and if $A$ is measurable subset of $X$, then $$\int_A f d\mu = \int_X f \chi_A d\mu.$$

Proof. If $s$ belong to $\Gamma_f$ then $s \chi_A \in \Gamma_{f\chi_A}$, so $\int_A s d\mu = \int_X s\chi_A f d\mu \le \int_X f\chi_A d\mu.$ Taking the supremum of all $s \in \Gamma_f$ we get $\int_A f d\mu \le \int_X f\chi_A d\mu$. Contrary, if $t\in \Gamma_{f\chi_A}$, then $0 \le t(1-\chi_A) \le f \chi_A (1-\chi_A) =0$, that is $t=t \chi_A \in \Gamma_f$, so $\int_X t d\mu = \int_X t \chi_A d\mu = \int_A t d\mu \le \int_A f d\mu.$ Taking the supremum from all $t\in \Gamma_{f\chi_A}$ we get $\int_X f \chi_A d\mu  \le \int_A f d\mu$.
Now, we are ready for definition of integral for complex function. First, we define $$\mathfrak{L}^1(\mu)= \left\{f:X \to \mathbb{C}: f \text{ is measurable and } \int_X |f| d\mu < + \infty\right \}.$$
The definition is correct because measurable of function $f$ implies measurable of function $|f|$.

Definition 3. Let $f=f_{\Re}+if_{\Im} \in \mathfrak{L}^1$, where $f_{\Re}=\mathfrak{Re} \circ f$ and $f_{\Im} = \mathfrak{Im}\circ f$ are real component of function $f$, and let $f^{\pm}_{\Re},f^{\pm}_{\Im}$ be positive and negative parts that function and let $A$ is measurable subset of $X$. Then we define $$\int_A f d\mu = \int_A f^{+}_{\Re} d\mu - \int_A f^{-}_{\Re} d\mu + i \int_A f^{+}_{\Im} d\mu - i \int_A f^{-}_{\Im} d\mu.$$

Notation: If $f:X \to \overline{\mathbb{R}}$ is measurable function, then we define $$f^+ = \max (f,0)\, \text{ and }\, f^-=-\min(f,0),$$ and we called it postive, that is negative part of real function $f$. Note that $f=f^+-f^-,|f|=f^++f^-$ and that $f^+(x) \ge 0$ and $f^-(x) \ge 0$ for all $x\in X$.

Definition 4. Function from $\mathfrak{L}^1(\mu)$ we call Lebesgue integrable (on $X$ with measure $\mu$), $\int_A f d\mu$ is Lebesgue integral of function $f \in \mathfrak{L}^1 (\mu)$ on measurable set $A$.

And now you can prove (I leave this for you) that if $f \in \mathfrak{L}^1 (\mu)$ then $$\int_A f d\mu = \int_X f \chi_A d\mu,$$ where $A$ is measurable subset of $X$.
A: Since $E=A\cup(E/A)$ then the Lebesgue integrat for $f$ on $A$ is $$\int_Ef.\chi_A=\int_Af.\chi_A + \int_{E/A}f.\chi_A $$ $$\implies\int_Ef.\chi_A=\int_Af$$
Maybe this is the right way to answer this question because I had to present the answer on the board and my professor didn't point out any mistakes.
