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Suppose for any $N\in\mathbb{N}$ and any finite sequence $\{x_i\}_{i=1}^N$ with each $x_i\in[0,1]$, we have $\sum_{i=1}^N f(x_i)\geq \sum_{i=1}^Ng(x_i)$. Is there any way to extend this to show that $\int_0^1f(x)dx \geq \int_0^1 g(x)dx$?

The way I was approaching it is to think about $[0,1]$ as a sample space with the uniform measure and approximate it by taking $N$ partitions $\{[0,y_1), [y_1,y_2),\ldots, [y_{N-1}, 1]\}$ of $[0,1]$ and think of each interval having a uniform distribution over $[y_{k-1}, y_{k})$ for $k\geq 1$. Then maybe under some continuity/boundedness conditions on $f$ and $g$, we can get the extension by starting from a selection of $x_k\in[y_{k-1}, y_{k}]$

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    Use $N=1$. What does that give you? – 2017-01-27
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    lol quasi's comment provides a much simpler answer to your question that works for any Lebesgue integrable function on [0,1]. And it avoids taking limits – 2017-01-27

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If we know $f$ and $g$ are Riemann integrable on $[0,1]$, then we know $$\int_{0}^{1}f=\lim_{n\to\infty}\frac{1}{n}\sum_{m=1}^{n}f(m/n)$$ and similarly for $g$. Then your result would follow by observing that $$\frac{1}{n}\sum_{m=1}^{n}f(m/n)\geq \frac{1}{n}\sum_{m=1}^{n}g(m/n)$$ and then taking the limit of both sides as $n\to\infty$.