Let $\lambda$ denote the Lebesgue measure on $(\mathbb{R}, \mathcal{B}(\mathbb{R}))$. Let $f:\mathbb{R}\to\mathbb{R}$ be an integrable function.
Exercise: Show that for all $-\infty < a,b <\infty$ we have, $$\int\limits_{[a,b]}fd\lambda = \int\limits_{(a,b)} fd\lambda.$$ Furthermore, show whether this is true if we replace $\lambda$ by any arbitrary measure on $\mathcal{B}(\mathbb{R})$.
What I've tried:
Since $f$ is integrable we know that $\int_{[a,b]} f^+d\lambda$ and $\int_{[a,b]} f^-d\lambda$ are finite. Pick a sequence of simple functions $(f_n^+)_{n\geq1}$ such that $0\leq f_n^+\uparrow f^+$. We have that $\int_{[a,b]}f^+ d\lambda = \lim\limits_{n\to\infty}\int_{[a,b]}f_n^+d\lambda.$ As $f_n^+$ is a simple function we know that $\int_{[a,b]}f_n^+d\lambda = \sum\limits_{j=1}^nx_j \cdot \lambda([a,b])$. We know that $\lambda([a,b]) = \lambda((a,b))$, so $\int_{[a,b]}f_n^+d\lambda = \sum\limits_{j=1}^n x_j\cdot \lambda((a,b)) = \int_{(a,b)}f_n^+d\lambda$. This proof is similar for $\int_{[a,b]}f_n^-d\lambda$ and combining $\int_{[a,b]}f_n^-d\lambda$ and $\int_{[a,b]}f_n^+d\lambda$ leads to $\int\limits_{[a,b]}fd\lambda = \int\limits_{(a,b)} fd\lambda.$
Questions:
Is my proof correct? If not; what am I doin wrong?
How do I show that the statement is not true for any arbitrary measure?
Thanks in advance!
Your proof is good. For the second question: take the delta measure $\delta_x$ defined by: \begin{align} \delta_x(A) = \mathbf{1}_{x\in A} \end{align} You can show that this is indeed a measure. Take $x=0$ so $\delta_0((0,1))=0$ and $\delta_0([0,1])=1$, and so $\delta_0((0,1))\neq \delta_0([0,1])$. Now just integrate the constant function $f(x)=1$ (why is this integrable?). \begin{align} 0=\int_{(0,1)} f\,d \delta_0 \neq \int_{[0,1]}f\, d\delta_0=1 \end{align}