# Find $\lim_{n \rightarrow \infty} \int_{[0,2]} t^{1/n} \, d\mu(t)$ with Dirac measure

I want to find the integral of: $$\lim_{n \rightarrow \infty} \int_{[0,2]} t^{1/n} \, d\mu(t)$$ where $$\mu=\delta_0 + \delta_1+ \delta_2$$ (Dirac measures)

My problem is that I can't determine if the integral is 2 or 3. Because for $$t=0$$ I have $$0^{1/n} \rightarrow 0$$ so $$t=0$$ does not "contribute" to the integral. However it is part of the measure as $$\mu([0,2])=1+1+1=3$$. Maybe I can say that it is 3 almost everywhere but not sure

Any help/hint would be appreciated

Directly from definitions- $$\int_{[0,2]} t^{1/n} \, d\mu(t) = 0^{1/n} + 1^{1/n} + 2^{1/n} = 1 + 2^{1/n} \xrightarrow[n\to\infty]{} 2$$
• How are you using the measure? To me it looks more like the integral of a indicator function like $\int_{[0,2]} t^{1/n} \cdot 1_{\{0\},\{1\},\{2\}} d\mu(t)$. Furthermore $\mu([0,2])=3$ and I don't see where you are using this – Daniel Nov 3 '19 at 7:07
• @Daniel 1. Every function is equal $\mu$-almost everywhere to a simple function of three terms $a1_{\{0\}}+ b 1_{\{1\}} + c1_{\{2\}}$ (possibly taking negative or complex values). You can write it as three integrals against 3 different Dirac measures if that helps. 2. where do you use that the Lebesgue measure of $[0,1]$ is 1 when computing usual Lebesgue measure integrals over $[0,1]$? The answer is that you don't – Calvin Khor Nov 3 '19 at 8:42