Proving Openness of {f(x)>a} in R

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"Let (f_n) be an increasing sequence of continuous functions on R. Suppose \forall x\in\mathbb{R}(f(x)=\lim_{n\rightarrow\infty}f_n(x)), and suppose that f(x)<\infty for all x, prove that \{x\in\mathbb{R}:f(x)>a\} is open for all a in R."

I think an additional condition of uniform convergence is required.
 
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No, you don't need uniform convergence. The proof is essentially a one- or two-liner.
 
I got it. Thanks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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