I want to show that if $(X, M,\mu)$ is a measure space, $A_1, \ldots, A_n, B_1, \ldots, B_n \in M$, $\alpha_1,\ldots, \alpha_n \in \mathbb{R}$,
$$f=\sum_{j=1}^n \alpha_j \chi_{A_j}, \ \ g=\sum_{j=1}^n \alpha_j \chi_{B_j} $$
and $f=g$ almost everywhere, then $\int f d\mu=\int g d\mu$.
Can you help me?. Thanks in advance.
On
Let $h = f-g$. Then $h\equiv 0$ a.e.
Define $A = \{x \in X : f(x) \neq g(x)\}$, i.e. the null-set, where $f$ and $g$ differ.
note, $|h| \leq 1_{A} \cdot \max\{\sup|f|, \sup|g|\}$.
$|h| \geq 0$ so we can try to calculate its integral, and it follows $$\int |h|\,\mathrm{d}\mu \leq \int 1_{A} \cdot \max\{\sup|f|, \sup|g|\}\, \mathrm{d} \mu= 0 \cdot \max\{\sup|f|, \sup|g|\} = 0$$
But $\int |h|\,\mathrm{d}\mu = \int h^+\,\mathrm{d}\mu + \int h^-\, \mathrm{d}\mu$, hence both of them are $0$.
$$\Rightarrow \int h\,\mathrm{d}\mu = \int h^+\,\mathrm{d}\mu - \int h^-\, \mathrm{d}\mu = 0$$
From linearity of Lebesgue integral you, on the other hand, have $$0 = \int h\,\mathrm{d}\mu = \int f\,\mathrm{d}\mu - \int g\,\mathrm{d}\mu$$
On
I am assuming that $f,g$ are integrable.
$h=f-g$ is simple, let $\beta_1,..,\beta_m$ be the non zero elements of the range of $h$, then $h = \sum_k \beta_k 1_{h^{-1}(\{\beta_k\})}$.
We are given that $\mu (\cup_k h^{-1}(\{\beta_k\}) ) = 0$, hence each $h^{-1}(\{\beta_k\})$ mas measure zero.
Then $\int h d\mu = \sum_k \beta_k \mu (h^{-1}(\{\beta_k\})) = 0$.
Let $h=f-g$. Then $h =\sum_1 ^{n} \alpha_j \chi_{ C_j} $ where $C_j =A_j \Delta B_j$, the symmetric difference between $A_j$ and $B_j$. Assuming that $\alpha_j$ are distinct (which is what is usually done in defining integrals of simple function) we get $A_j =f^{-1} \{\alpha_j\}$ and $B_j =g^{-1} \{\alpha_j\}$ so $C_j \subset \{x ;f(x) \neq g(x)$. Hence $\mu (C_j)=0$ for each j which gives $\int h d\mu =0$ or $\int f d\mu =\int g d\mu$. The result is true even if $\alpha_j$'s are not distinct but the proof is messier.