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Let f∈L1(R,B(R),μ)f\in\mathscr{L}^{1}(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu)f∈L1(R,B(R),μ), let φ:R→R\varphi:\mathbb{R}\to \mathbb{R}φ:R→R convex (implying continuity, implying measurability) and assume φ∘f∈L1(R,B(R),μ)\varphi\circ f\in\mathscr{L}^{1}(\mathbb{R},\mathcal{B}(\mathbb{R}),\mu)φ∘f∈L1(R,B(R),μ). Also, μ(R)=1\mu(\mathbb{R})=1μ(R)=1 then φ(∫Rf dμ)≤∫R(φ∘f) dμ\varphi\left( \int\limits _{\mathbb{R}}f \, d\mu \right)\le \int\limits _{\mathbb{R}}(\varphi\circ f) \, d\mu φR∫fdμ≤R∫(φ∘f)dμ ## Corollary ∣∫Rf dμ∣≤∫R∣f∣ dμ\left|\int\limits _{\mathbb{R}}f \, d\mu \right|\le \int\limits _{\mathbb{R}}|f| \, d\mu R∫fdμ≤R∫∣f∣dμ ## Example exe^{x}ex is Convex on R\mathbb{R}R so we have for any f∈L1(μ)f\in\mathscr{L}^{1}(\mu)f∈L1(μ):e∫f dμ≤∫ef dμe^{\int\limits f \, d\mu }\le\int\limits e^{f} \, d\mu e∫fdμ≤∫efdμ