Is $\int g\, d\mathbb{P}=\int g\, d\mathbb{P}_{|\mathfrak{F}}$? 

Let $(\Omega,\mathfrak{A},\mathbb{P})$ be a probability space and $\mathfrak{F}\subset\mathfrak{A}$ a sub-$\sigma$-algebra. Consider $g\in L_{\mathbb{P}}^1$ and $g$ $\mathcal{F}$-measurable. Let $\mathbb{P}_{|\mathfrak{F}}$ be the restriction from $\mathbb{P}$ on $\mathfrak{F}$. My question is if then
    $$
\int g\, d\mathbb{P}=\int g\, d\mathbb{P}_{|\mathfrak{F}}.
$$


Can you give me a hint how to prove or disprove that? I did not have an idea yet.
 A: First I have to show that for elementar function. So let
$$
g=\sum_{i=1}^n\alpha_i\chi_{A_i}, A_i:=\left\{g=\alpha_i\right\}, A_i\in\mathfrak{F}, \Omega=\biguplus_{i=1}^{n}A_i,
$$
then
$$
\int g\, d\mathcal{P}_{|\mathfrak{F}}=\sum_{i=1}^{n}\alpha_i\mathbb{P}_{|\mathfrak{F}}(A_i).
$$
Now, because of $\mathfrak{F}\subset\mathfrak{A}$ it is $A_i\in\mathfrak{A}, i=1,\ldots,n$. So it is
$$
\int g\, d\mathbb{P}=\sum_{i=1}^{n}\alpha_i\mathbb{P}(A_i).
$$
Because $A_i\in\mathfrak{F}, i=1,\ldots,n$ it is $\mathbb{P}_{|\mathfrak{F}}(A_i)=\mathbb{P}(A_i), i=1,\ldots,n$, so the both sums are equal and so the integrals.
Now for $g\geq 0$, g measurable. Then it exists a series $(g_n)$ of elementar functions with $g_n\nearrow g$. So with what I've already shown for elementar functions and Beppo Levi resp. monotone convergence, it is
$$
\int g\, d\mathbb{P}=\lim_n\int g_n\, d\mathbb{P}=\lim_n\int g_n\, d\mathbb{P}_{|\mathfrak{F}}=\int g\, d\mathbb{P}_{|\mathfrak{F}}.
$$
Finally for any measurable function $g$. First of all $g$ is integrable, so for $g=g^+-g^-$ it is by what is alredy shown
$$
\infty>\int g^+\, d\mathbb{P}=\int g^+\, d\mathbb{P}_{|\mathfrak{F}},~~\infty> \int g^-\, d\mathbb{P}=\int g^-\, d\mathbb{P}_{|\mathfrak{F}}.
$$
So it is
$$
\int g\, d\mathbb{P}=\int g^+\, d\mathbb{P}-\int g^-\, d\mathbb{P}=\int g^+\, d\mathbb{P}_{|\mathfrak{F}}-\int g^-\, d\mathbb{P}_{|\mathfrak{F}}=\int g\, d\mathbb{P}_{|\mathfrak{F}}.
$$
