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Just out of curiosity, and this may be something trivial, but can any function $f:\mathbb{R} \rightarrow \mathbb{R}_{+} \cup \{0\}$ (other than $f\equiv 0$) give us the following inequality:

$$\forall c

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    You would need this to be true in the limit $d\downarrow b$ and $c\uparrow a$ for any $a$ and $b.$ But that can only work if $b-a = 1$ (Unless you meant to put a $1/(c-d)$ out front of the right hand integral) – 2017-02-17
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    For your example. Fix $a,b,c$, and $d$; if $f = 1\{x \in [a,b]\}$ then the inequaily fails. – 2017-02-17

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The function

$$s\mapsto \int_a^s f(x) dx$$

is continuous in $s$ (and similar for the lower bound). So let $a\to c, b\to d$, we have

$$ \frac{1}{d-c} \int_c^d f(x) dx \le \int_c^d f(x) dx.$$

Since $c, d$ are arbitrary, let $d = c+ 1/2$ gives

$$2 \int_c^{c+1/2} f(x) dx \le \int_c^{c+1/2} f(x)dx$$

Since $f$ is nonnegative, this gives

$$\int_c^{c+1/2} f(x) dx = 0$$

for all $c$. This implies that $f$ is zero "almost everywhere". (there is a precise definition for this notation, when you learn measure theory). Some examples are

$$ f(x) = \begin{cases} 1 & \text{if } x = 1,1/2,1/3,\cdots \\ 0 &\text{otherwise.}\end{cases}$$

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    Very nice argument (+1)! As a side note, without referencing to any measure theory, Johnsonbaugh and Pfaffenberger, in their book *Foundations of Mathematical Analysis*, devote section 57 to sets of measure zero on the real line and proceed to define "almost everywhere" in that context. They do so in order to prove, in the next section, that a function $f$ is Riemann integrable on $[a,b]$ if and only if $f$ is continuous almost everywhere in $[a,b]$. As their Corollary 58.6, they show that if $f$ is Riemann integrable on $[a,b]$ and $\int_a^b |f|=0$ then $f(x)=0$ almost everywhere on $[a,b]$. – 2017-02-17