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Let $C_{00}(\mathbb{R})$ denote the set of all continuous functions $f:\mathbb{R}\rightarrow \mathbb{C}$ with compact support. Let $\alpha$ and $\beta$ be continuous, real-valued, nondecreasing functions on $\mathbb{R}$ such that $\alpha(0) = \beta(0) = 0 $ and $\alpha\neq\beta$. I need to find a positive valued function in $C_{00}(\mathbb{R})$ such that $$\int f\ d\alpha\neq\int f\ d\beta.$$ Please help.

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Let $t\in \mathbb{R}$ be a point such that $\alpha(t)\neq \beta(t)$. If we define $f=\chi_{[0,t]}$ then $$\int fd\alpha=\alpha(t)\neq\beta(t)=\int f d\beta.$$ Now let us define $\ f_n(x)=1\ for\ x\in [0,t]$, $\ f_n$ $is\ linear\ in\ the\ intervals$ $[-\frac{1}{n},0]$ $and$ $[t,t+\frac{1}{n}]$, $and\ it\ is\ zero\ on\ the\ rest\ of\ \mathbb{R}$ $so\ that$ $f_n\in C_{00}(\mathbb{R})$. Since $\alpha$ is uniformly continuous on $[-1,t+1]$, it can be shown that $$\int f_n\ d\alpha\rightarrow \int f\ d\alpha.$$ Actually $$|\int f_n\ d\alpha - \int f\ d\alpha|\leq\alpha(0)-\alpha(-\frac{1}{n})+\alpha(t+\frac{1}{n})-\alpha(t)$$ and hence the above follows. Similar argument shows that $$\int f_n\ d\beta\rightarrow \int f\ d\beta.$$ Since $$\int fd\alpha=\alpha(t)\neq\beta(t)=\int f d\beta,$$ for sufficiently large $n$ we have $$\int f_{n}\ d\alpha\neq\int f_{n}\ d\beta.$$

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